Test Questions on Chapter # 1 ………….. Stoichiometry (Test # 2)
IMPORTANT NUMERICALS on Stoichiometry from BOOK
Q1.Calculate the mass of carbon dioxide (CO2) that can be obtained by complete thermal decomposition of 50 g of lime stone (CaCO3)
CaCO3(s) → CaO(s) + CO2(g)
(Answer; 22 g)
Q2.Mass of 49 g of solid potassium chlorate (KClO3) on heating decomposes completely to potassium chloride (KCl) with the liberation of oxygen gas(O2)
2KClO3 → 2KCl + 3O2
Determine volume of oxygen gas (O2) liberated at STP.
(Answer; 13.44 dm3)
Q3.Silver sulphide (Ag2S) is an anti microbial agent. In an experiment, 24.8g Ag2S is reacted with the excess of hydrochloric acid as given in the following reaction:
Ag2S(s) + 2HCl(aq) → 2AgCl(s) + H2S(g)
Calculate the
(i)Mass of AgCl formed [Answer; 28.7 g]
(ii)Volume of H2S produced at STP[Answer; 2.24 dm3]
(Atomic mass of Ag is 108 and S is 32)
Q4.On heating solid ammonium nitrate (NH4NO3), it decomposes to produce nitrous oxide (N2O) and water
NH4NO3(s) → N2O(g) + 2H2O(l)
If 200 g of ammonium nitrate is completely consumed in the reaction, calculate the:
(i) Mass of water formed
(ii) Volume of N2O gas liberated at STP
(Answer; 90 g water, 56 dm3 of O2)
Q5. Calculate the volume of carbon dioxide at STP that can be produced by the complete burning of 50 dm3 of butane gas (C4H10) in the excess of supply of oxygen gas (O2).
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
(Answer; 200 dm3)
IMPORTANT NUMERICALS on Stoichiometry from External Source
Q1. Calculate the volume of CO2 gas produced at standard temperature and pressure by combustion of 50 g of methane.
CH4 + 2O2 → CO2 + 2H2O
(Ans; 70 dm3)
Q2. Calculate the volume of O2 at S.T.P. produced by the complete decomposition of 540 g of N2O5
2N2O5 → 4NO2 + O2
(Ans; 56 dm3)
Q3. 100 grams of KNO3 are heated to redness. What volume of oxygen at 39°C and 765 mm pressure will evolve?
(Karachi Board-2006)
2KNO3 → 2KNO2 + O2
(Ans; 11.089 dm3)
Q4. Potassium chlorate if often used to generate oxygen gas in high school laboratory. If 183.7 g of KClO3 is completely burnt catalytically, what volume of oxygen gas will be obtained at 39○C and 1200 torr pressure?
Q5. Calculate the volume of nitrogen gas produced by heating 800 g of ammonia at 21°C and 823 torr pressure?
2NH3→ N2+ 3H2 (KB- 2014)
(Ans; 522.19 dm3)
Q6. CaCO3 is often used to generate CO2 gas in industry. If 200 g of CaCO3 is strongly heated, what volume of CO2 gas will be obtained at 30°C and 1200 torr pressure? (KB- 2017)
CaCO3(s) → CaO(s) + CO2(g)
(Ans; 31. 49 dm3)
Q7. Zinc reacts with H2SO4 (dil) as given below:
Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)
Calculate the mass of ZnSO4, the volume of H2 gas at STP and the number of molecules of H2 gas which will be produced by reacting 163.5 g of Zn with H2SO4.
(Ans;403.5 g of ZnSO4, 56 dm3 of H2 gas, 1.505 x 1024 molecules of H2 gas)
IMPORTANT NUMERICALS on Limiting Reactant from BOOK
Q1. How many grams of NH3 are formed when 100 g of each of the following reagents are reacted together according to following equation;
2NH4Cl(s) + Ca(OH)2(s) → CaCl2(s) + 2NH3(g) + 2H2O(l)
(Ans; 31.62 g)
Q2. 5.6 g of NH3 are allowed to react with 4.5 g of O2 according to following equation. Calculate number of moles, number of molecules, mass in gram and volume in cm3 of nitric oxide evolved at STP.
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
(Ans; 0.1128 mol NO, 6.79 x 1023 molecules of NO, 3.384 g NO, 2526.72 cm3 NO)
Q3. Combustion of ethene in air to form CO2 and H2O is given in the following equation
C2H4(g) +3O2(g) → 2CO2(g) + 2H2O(l)
If a mixture containing 2.8 g C2H4 and 6.4 g O2 is allowed to ignite, identify the limiting reactant and determine the mass of CO2 gas will be formed.
(Answer; 5.852 g)
Q4. Ammonia gas can be produced by heating together solid NH4Cl and Ca(OH)2
2NH4Cl + Ca(OH)2 → CaCl2 + 2NH3 + 2H2O
If a mixture containing 100 g of each these solids is heated, determine the limiting reactant and mass of NH3 gas produced
[Answer; Limiting reactant is NH4Cl, Mass of NH3 is 31.7 g]
Q5. When aluminium is heated with nitrogen at 700oC, it gives aluminium nitride.
2Al(s) + N2(g) → 2AlN(s)
If 67.5 g of aluminium and 140 g of nitrogen gas are allowed to react, find out:
(a) Limiting reactant (Answer; AlN)
(b)Mass of aluminum nitride produced?
(Answer; 102.5 g AlN
(c) Mass of excess reactant? (Answer; 105 g N2)
Numericals on Stoichiometry (Mass-Mass, Mass-Volume, Vol-Vol relationships)
Q1. Calculate the mass of carbon dioxide (CO2) that can be obtained by complete thermal decomposition of 50 g of lime stone (CaCO3)
CaCO3(s) → CaO(s) + CO2(g)
Solution
Calculation of moles of CaCO3
Mass of CaCO3 = 50 g
Molar mass of CaCO3 = 40 + 12 + 3(16) = 100 gmol−1
Molar mass of CO2 = 12 + 2(16) = 44 gmol−1
Mass of CO2 = ?
No. of moles of CaCO3 = mass/molar mass = 50/100 = 0.5 mol
Mole Comparison of CaCO3 and CO2
CaCO3(s) → CaO(s) + CO2(g)
(Given) (Required)
1 mol----------------> 1 mol
0.5 mol -------------> ?
According to unitary method
1 moles of CaCO3 produces 1 mole CO2
0.5 mole of CaCO3 produces 0.5 mole CO2
Calculation of moles of CO2
Mass of CO2 = n x M = 0.5 x 44 = 22 g
Q2. Mass of 49 g of solid potassium chlorate (KClO3) on heating decomposes completely to potassium chloride (KCl) with the liberation of oxygen gas(O2)
2KClO3 → 2KCl + 3O2
Determine volume of oxygen gas (O2) liberated at STP.
Solution
Calculation of moles of KClO3
Mass of KClO3 = 49 g
Molar mass of KClO3 = 39 + 35.5 + 3(16) = 122.5 gmol−1
Molar mass of O2 = 2(16) = 32 gmol−1
Volume of O2 at STP = ?
No. of moles of KClO3 = mass/molar mass = 49/122.5 = 0.4 mol
Mole Comparison of KClO3 and O2
2KClO3 → 2KCl + 3O2
(Given) (Required)
2 mol --------------------> 3mol
0.4 mol ------------------> ?
According to unitary method
2 moles of KClO3 produces 3 mole O2
1 moles of KClO3 produces 3/2 mole O2
0.4 mole of KClO3 produces 3/2 x 0.4 mole O2 = 0.6 mol O2
Calculation of Volume O2 at STP
Volume of O2 at STP = n x molar volume = 0.6 x 22.4 = 13.44 dm3
Q3. On heating solid ammonium nitrate (NH4NO3), it decomposes to produced nitrous oxide (N2O) and water.
NH4NO3(s) → N2O(g) + 2H2O(l)
If 200 g of ammonium nitrate is completely consumed in the reaction, calculate the
(i) Mass of water formed
(ii) Volume of N2O gas liberated at STP
Solution
Calculation of moles of NH4NO3
Mass of NH4NO3 = 200 g
Molar mass of NH4NO3 = 14 + 4(1) + 14 + 3(16) = 80 gmol−1
Molar mass of H2O = 2(1) + 16 = 18 gmol−1
Molar volume at STP = 22.4 dm3
Mass of water formed = ?
Volume of N2O at STP = ?
No. of moles of NH4NO3 = mass/molar mass = 200/80 = 2.5 mol
Mole Comparison of NH4NO3 and N2O
NH4NO3(s) → N2O(g) + 2H2O(l)
(Given) (Required)
1 mol -----------------------> 2 mol
2.5 mol ---------------------> ?
According to unitary method
1 moles of NH4NO3 produces 2 mole H2O
1 moles of NH4NO3 produces 2/1mole H2O
2.5 mole of NH4NO3 produces 2 x 2.5 mole H2O = 5 mol H2O
Calculation of Mass of H2O
Mass = n x molar mass = 5 x 18 = 90 g H2O
Calculation of Volume O2 at STP
Mole Comparison of NH4NO3 and N2O
NH4NO3(s) → N2O(g) + 2H2O(l)
(Given) (Required)
1 mol -----------> 1mol
2.5 mol ---------> ?
According to unitary method
1 moles of NH4NO3 produces 1 mole N2O
2.5 mole of NH4NO3 produces 1 x 2.5 mole N2O = 2.5 mol N2O
Volume of N2O at STP = n x molar volume = 2.5 x 22.4 = 56 dm3 N2O
Q4. Calculate the volume of carbon dioxide at STP that can be produced by the complete burning of 50 dm3 of butane gas (C4H10) in the excess supply of oxygen gas (O2).
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
Solution
Volume of butane at STP = 50 dm3
Volume of CO2 at STP = ?
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
(Given) (Required)
2 mol 8 mol
2 Volume 8 volume (applying Gay-Lussac’s Law)
2 dm3 --------------------> 8 dm3
50 dm3 ------------------> ?
According to unitary method from balanced chemical equation
2 dm3 of butane gives = 8 dm3 of carbon dioxide at STP
2 dm3 of butane gives = 8/2 dm3 of carbon dioxide at STP
50 dm3 of butane gives = 8/2 x 50 = 200 dm3 of carbon dioxide at STP
Q5. Silver sulphide (Ag2S) is an anti microbial agent. In an experiment, 24.8g Ag2S is reacted with the excess of hydrochloric acid as given in the following reaction: (Atomic mass; Ag =108 and S = 32)
Ag2S(s) + 2HCl(aq) → 2AgCl(s) + H2S(g)
Calculate the
(i) Mass of AgCl formed
(ii)Volume of H2S produced at STP
Solution
Calculation of moles of Ag2S
Mass of Ag2S = 24.8 g
Molar mass of Ag2S = 2(108) + 32 = 248 gmol−1
Molar mass of AgCl = 108 + 35.5 = 143.5 gmol−1
Molar volume at STP = 22.4 dm3
Mass of AgCl formed = ?
Volume of H2S at STP = ?
No. of moles of NH4NO3 = mass/molar mass = 24.8/248 = 0.1 mol
Mole Comparison of Ag2S and AgCl
Ag2S(s) + 2HCl(aq) → 2AgCl(s) + H2S(g)
(Given) (Required)
1 mol -----------------> 2mol
0.1 mol ---------------> ?
According to unitary method
1 moles of Ag2S produces 2 mole AgCl
0.1 mole of Ag2S produces 2 x 0.1 mole H2O = 0.2 mol AgCl
Calculation of Mass of AgCl
Mass = n x molar mass = 0.2 x 143.5 = 28.7 g AgCl
Calculation of Volume H2S at STP
Mole Comparison of Ag2S and H2S
Ag2S(s) + 2HCl(aq) → 2AgCl(s) + H2S(g)
(Given) (Required)
1 mol --------------------------> 1 mol
0.1 mol ------------------------> ?
According to unitary method
1 moles of Ag2S produces 1 mole H2S
0.1 mole of Ag2S produces 1 x 0.1 mole H2O = 0.1 mol H2S
Volume of H2S at STP = n x molar volume = 0.1 x 22.4 = 2.24 dm3 H2S
Numericals on Limiting Reactant
Q1.Combustion of ethene in air to form CO2 and H2O is given in the following equation
C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l)
If a mixture containing 2.8 g C2H4 and 6.4 g O2 is allowed to ignite, identify the limiting reactant and determine the mass of CO2 gas will be formed.
Solution
(I) Conversion of Mass of Each Reactant into Moles
Mass of C2H4 (ethene) = 2.8 g
Mass of O2 = 6.4 g
Molar mass of C2H4 = 2(12) + 4(1) = 28 g/mol
Molar mass of O2 = 2(16) = 32 g/mol
No. of moles (n) of C2H4 = mass/molar mass = 2.8/28 = 0.1 mol
No. of moles (n) of O2 = mass/molar mass = 6.4/32 = 0.2 mol
(II) Calculation of molar amount of product from molar amount of each reactant
C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l)
(Given) (Given) (Required)
1 mole 2 mole ---------> 2 mole
0.1 mole 0.2 mole -------> ?
Mole Comparison of C2H4 and CO2
1 moles of C2H4 produces 2 mole CO2
0.1 mole of C2H4 produces 2 x 0.1 mole CO2 = 0.2 mole CO2
Mole Comparison of C2H4 and O2
3moles of O2 produces 2 mole CO2
1 mole of NH4Cl produces 2/3 mole CO2
0.2 mole of NH4Cl produces 2/3 x 0.2 mole CO2 = 0.133 mole CO2
(
From above calculation, it is clear that least amount of product (CO2) is produced by O2 so O2 is a limiting reactant and amount of product is calculated from its amount.
(IV) Conversion of Molar Amount of Product into Gram
Molar amount of CO2 = 0.133 moles
Molar mass of CO2= 12 + 2(16) = 44 g/mol
Amount of CO2 in g = moles x molar mass = 0.133 x 44 = 5.852 g CO2
Q2.How many gram of NH3 are formed when 100 g of each of the following reagents are reacted together according to following equation:
2NH4Cl + Ca(OH)2 → CaCl2 + 2NH3 + 2H2O
Solution
(I) Conversion of Mass of Each Reactant into Moles
Mass of NH4Cl = 100 g
Mass of Ca(OH)2 = 100 g
Molar mass of NH4Cl = 14 + 4 + 35.5 = 53.5 g/mol
Molar mass of Ca(OH)2= 40 + 32 + 2 = 74 g/mol
No. of moles (n) of NH4Cl = mass/molar mass = 100/53.5 = 1.869 mol
No. of moles (n) of Ca(OH)2 = mass/molar mass = 100/74 = 1.351mol
(II) Calculation of molar amount of product from molar amount of each reactant
2NH4Cl + Ca(OH)2 → CaCl2 + 2NH3 + 2H2O
(Given) (Given) (Required)
2 mole 1 mole -----------> 2 mole
1.869 mole 1.315 mole -----------> ?
Mole Comparison of NH4Cl and NH3
2 moles of NH4Cl produces 2 mole NH3
1 mole of NH4Cl produces 2/2 mole NH3
1.869 mole of NH4Cl produces 2/2 x 1.869 mole NH3 = 1.869 mole NH3
Mole Comparison of Ca(OH)2 and NH3
1 mole of Ca(OH)2 produces 2 mole NH3
1.351 mole of Ca(OH)2 produces 2/1 x 1.351 mole NH3 = 2.7 mole NH3
(
From above calculation, it is clear that least amount of product (NH3) is produced by NH4Cl so NH4Cl is a limiting reactant and amount of product is calculated from its amount.
(IV) Conversion of Molar Amount of Product into Gram
Molar amount of NH3= 1.869 moles
Molar mass of NH3 = 14 + 3(1) = 17 g/mol
Amount of NH3 in g = moles x molar mass =1.869 x 17 = 31.773 g NH3
Q3.When aluminium is heated with nitrogen at 700oC, it gives aluminium nitride.
2Al(s) + N2(g) → 2AlN(s)
If 67.5 g of aluminium and 140 g of nitrogen gas are allowed to react, find out:
(a) Limiting reactant
(b) Mass of aluminum nitride produced?
(c) Mass of excess reactant?
Solution
(I) Conversion of Mass of Each Reactant into Moles
Mass of Al = 67.5 g
Mass of N2 = 140 g
Molar mass of Al = 27 g/mol
Molar mass of N2 = 2(14) = 28 g/mol
No. of moles (n) of NH4Cl = mass/molar mass = 67.5/27 = 2.5 mol
No. of moles (n) of Ca(OH)2 = mass/molar mass = 140/28 = 5.0 mol
(II) Calculation of molar amount of product from molar amount of each reactant
2Al(s) + N2(g) → 2AlN(s)
(Given) (Given) (Requreid)
2 mole 1 mole ----> 2 mole
2.5 mole 5.0 mole ---> ?
Mole Comparison of Al and AlN
2 moles of Al produces 2 mole AlN
1 mole of Al produces 2/2 mole AlN
2.5 mole of Al produces 2/2 x 2.5 mole AlN = 2.5 mole AlN
Mole Comparison of N2 and AlN
1 moles of N2 produces 2 mole AlN
5.0 mole of N2 produces 2 x 5.0 mole AlN = 10 mole AlN
(
From above calculation, it is clear that least amount of product (AlN) is produced by Al so Al is a limiting reactant and amount of product is calculated from its amount.
(IV) Conversion of Molar Amount of Product into Gram
Molar amount of AlN = 2.5 moles
Molar mass of AlN = 27 + 14 = 41 g/mol
Amount of AlN in g = moles x molar mass = 2.5 x 41 = 102.5 g AlN
(V) Calculation of Mass of Excess Reactant
2Al(s) + N2(g) → 2AlN(s)
(Given) (Requreid)
2 mole ------> 1 mole
2.5 mole -----> ?
Mole Comparison of Al and N2
2 moles of Al reacts with 1 mole N2
1 mole of Al produces ½ mole N2
2.5 mole of Al produces ½ x 2.5 mole N2 = 1.25 mole N2
Excess moles of N2 = Moles used – mole consumed = 5 – 1.25 = 3.75 moles N2 left behind
Conversion of excess mole into mass
Excess amount or mass of N2 = n x molar mass = 3.75 x 2(14) = 105 g N2 left behind
Q4.Hydrogen gas is commercially prepared by steam methane process
CH4 + H2O → CO + 3H2
If a mixture of 28.8 g methane and 14.4 g steam is heated in a furnace at elevated temperature to liberate carbon monoxide and hydrogen gas, determine the limiting reactant and the mass of hydrogen gas produced.
Solution
(I) Conversion of Mass of Each Reactant into Moles
Mass of CH4 = 28.8 g
Mass of H2O = 14.4 g
Molar mass of CH4 = 12+4(1) =16 g/mol
Molar mass of H2O = 2(1)+16 = 18 g/mol
No. of moles (n) of CH4 = mass/molar mass = 28.8/16 = 1.8 mol
No. of moles (n) of H2O = mass/molar mass = 14.4/18 = 0.8 mol
(II) Calculation of molar amount of product from molar amount of each reactant
CH4 + H2O → CO + 3H2
(Given) (Given) (Required)
1 mole 1 mole ------------> 3mole
1.8 mole 0.8 mole ----------> ?
Mole Comparison of CH4 and H2
1 moles of CH4 produces 3 mole H2
1.8 mole of CH4 produces 3 x 1.8 mole H2 = 5.4 mole H2
Mole Comparison of H2O and H2
1 moles of H2O produces 3 mole H2
0.8 mole of H2O produces 3 x 0.8 mole H2 = 2.4 mole H2
(
From above calculation, it is clear that least amount of product (H2) is produced by H2O so H2O is a limiting reactant and amount of product is calculated from its amount.
(IV)Conversion of Molar Amount of Product into Gram
Molar amount of H2=2.4 moles
Molar mass of H2 = 2(1) = 2 g/mol
Amount of H2 in g = moles x molar mass = 2.4 x 2 = 4.8 g H2
Q5. How many grams of Na2S2O3 will be produced when 200 g each of the three reagents are reacted together according to following equation?
2Na2S + Na2CO3 + 4SO2 → 3Na2S2O3 + CO2
Solution
1. Determination of molar amounts of reactants
Mass of each reactant = 200 g
Molar mass of Na2S = 46 + 32 = 78 g/mol
Molar mass of Na2CO3 = 46 + 12 + 48 = 106 g/mol
Molar mass of SO2 = 32 + 32 = 64 g/mol
No. of moles of Na2S in 200 g = mass/molar mass = 200/78 = 2.56 moles
No. of moles of Na2CO3 in 200 g = mass/molar mass = 200/106 = 1.88 moles
No. of moles of SO2 in 200 g = mass/molar mass = 200/64 = 3.12 moles
2. Calculation of molar amount of products from molar amount of each reactant
2Na2S + Na2CO3 + 4SO2 → 3Na2S2O3 + CO2
(Given) (Given) (Given) (Required)
2 mole 1 mole 4 mole -----------> 3 mole
2.56 mol 1.88 mol 3.12 mol --------> ?
2 moles of Na2S produces 3 moles Na2S2O3
1 mole of Na2S produces 3/2 moles Na2S2O3
.56 moles of Na2S produces 3/2 x 2.56 moles Na2S2O3 = 3.84 moles Na2S2O3
1 mole of Na2CO3 produces 3 moles Na2S2O3
1.88 mole of Na2CO3 produces 3 x 1.88 moles Na2S2O3 = 5.64 moles Na2S2O3
4 moles of SO2 produces 3 moles Na2S2O3
1 mole of SO2 produces ¾ moles Na2S2O3
3.12 mole of SO2 produces ¾ x 3.12 moles Na2S2O3 = 2.34 moles Na2S2O3
3. Determination of L.R. and Mass of Product
From above calculation, it is clear that least amount of product (Na2S2O3) is produced by SO2, so SO2 is a limiting reactants. So amount of SO2 is used to calculate amount of Na2S2O3.
Molar amount of Na2S2O3 = 2.34 moles
Molar Mass of Na2S2O3 = 2(23) + 2(32) + 3(16) = 46 + 64 + 48 = 158 g/mole
Amount of Na2S2O3 in g = moles x molar mass = 2.34 x 158 = 369.72 g Na2S2O3 (Answer)
Q6. 5.6 g of NH3 are allowed to react with 4.5 g of O2 according to following equation. Calculate number of moles, number of molecules, mass in gram and volume in cm3 of nitric oxide evolved at STP.
4NH3 +5O2 → 4NO + 6H2O
Solution
(I) Conversion of mass of each reactant into moles
Mass of NH3 = 5.6 g
Mass of O2 = 4.5 g
Molar mass of NH3 = 14+ 3(1) = 17 g/mole
Molar mass of O2 = 16 + 16 = 32 g/mole
No. of moles of NH3 = mass/molar mass = 5.6/17 = 0.3294 mol
No. of moles of O2 = mass/molar mass = 4.5/32 = 0.141 mol
(II) Calculation of molar amount of product from molar amount of each reactant
4NH3 + 5O2 → 4NO + 6H2O
(Given) (Given) (Required)
4 mole 5 mole ------------> 4 mole
0.3294 mol 0.141 mol--------> ?
4 moles of NH3 produces 4 moles of NO
1 mole of NH3 produces 4/4 mole of NO
0.3294 mol of NH3 produces 4/4 x 0.3294 mole of NO = = 0.3294 mol NO
5 moles of O2 produces 4 moles of NO
1 mole of O2 produces 4/5 moles of NO
0.141 mol of O2 produces 4/5 x 0.141 moles of NO = = 0.1128 mol NO
(
From above calculation, it is clear that least amount of product (NO) is produced by O2 so O2 is a limiting reactant and amount of product is calculated from its amount. Hence number of moles of NO formed will be 0.1128 mol
(IV) Conversion of Molar Amount of Product into Number of molecules
Number of molecules = n x NA ⇒ 0.1128 x 6.02 x 1023 = 6.79 x 1023 molecules of NO
(V) Conversion of Molar Amount of Product into Gram
Molar amount of NO = 0.1128 moles
Molar mass of NO = 14 + 16 = 30 g/mole
Amount of NO in g = moles x molar mass ⇒ 0.1128 x 30 = 3.384 g NO
(VI) Conversion of Molar Amount of Product into Volume
Volume of gas in cm3 = n x molar volume ⇒ 0.1128 x 22400 = 2526.72 cm3 NO
Q8.Calculate each of the following quantities
(ii) Mass in gram of 5.6 dm3 of ethene (C2H4) at STP
(ii) Number of molecules in 1 liter g of SO3
(iii) Volume in cm3 of 0.71 g Cl2 gas at STP
(iv) Number of formula units in 585 g of NaCl
Solution
Calculation of mass in gram of 5.6 dm3 of ethene (C2H4) at STP
Volume of ethene gas (C2H4) at STP = 5.6 dm3
Molar volume at STP = 22.4 dm3
No. of moles of given volume of ethene = ?
No. of moles (n) = Volume of gas at STP/Molar volume = 5.6/22.4 = 0.25 mol
Calculation of Number of molecules in 1 liter g of SO3
Volume of SO3at STP = 1 liter
Molar volume at STP = 22.4 liter
Avogadro’s number = 6.02 x 1023 molecules/mol
No. of molecules of given volume of SO3 = ?
No. of molecules = V at STP/molar volume x NA = 1/22.4 x 6.02 x 1023 = 2.69 x 1022 molecules
Calculation of Volume in cm3 of 0.71 g Cl2 gas at STP
Mass of Cl2 gas at STP = 0.71 g
Molar mass of Cl2 gas = 2(35.5) = 71 g mol−1
Molar volume at STP = 22.4 dm3
No. of molecules = mass/molar mass x molar volume = 0.71/71 x 22.4 = 0.224 dm3
Calculation of Number of formula units in 585 g of NaCl
Mass of NaCl = 585 g
Molar mass of NaCl = 23 + 35.5 = 58.5 g mol−1
Avogadro’s number = 6.02 x 1023 formula units/mol
No. of molecules = mass/molar mass x NA = 585/58.5 x 6.02 x 1023 = 6.02 x 1024 formula units
Solution of Chemistry Paper-A for XI
Q1.Calculate each of the following quantities
(i) Mass in gram of 3.01 x 1023 molecules of ethyne (C2H2)
(ii) Number of molecules in 6.4 g of SO2
(iii) Volume in cm3 of 38.4 g O2 gas at STP
(iv) Number of formula units in 333 g of CaCl2
Solution
Mass in gram of 3.01 x 1023 molecules of ethyne (C2H2)
No. of molecules of ethyne (C2H2) = 3.01 x 1023 molecules
Avogadro’s number = 6.02 x 1023 formula units/mol
Molar mass of ethyne (C2H2) = 2(12) + 2(1) = 26 g mol−1
Mass of ethyne (C2H2) = ?
Mass from particles = (Np/NA) x M = (3.01 x 1023 /6.02 x 1023) x 26 = 13 g
Number of molecules in 6.4 g of SO2
Mass of SO2 at STP = 6.4 g
Molar mass of SO2= 32 + 2(16) = 64 g mol−1
Avogadro’s number = 6.02 x 1023 molecules/mol
No. of molecules of given mass of SO2 = ?
No. of molecules = mass/molar mass x NA = 6.4/64 x 6.02 x 1023 = 6.02 x 1022 molecules
Volume in cm3 of 38.4 g O2 gas at STP
Mass of O2 gas at STP = 38.4 g
Molar mass of O2 gas = 2(16) = 32 g mol−1
Molar volume at STP = 22400 cm3
No. of molecules = mass/molar mass x molar volume = 38.4 /32 x 22400 = 26880 m3
Number of formula units in 333 g of CaCl2
Mass of CaCl2 = 333 g
Molar mass of CaCl2 = 40 + 2(35.5) = 111 g mol−1
Avogadro’s number = 6.02 x 1023 formula units/mol
No. of molecules = mass/molar mass x NA = 333/111 x 6.02 x 1023 = 1.806 x 1024 formula units
Q2. Mass of 49 g of solid potassium chlorate (KClO3) on heating decomposes completely to potassium chloride (KCl) with the liberation of oxygen gas (O2) (5)
2KClO3 → 2KCl + 3O2
Determine volume of oxygen gas (O2) liberated at STP.
Solution
Calculation of moles of KClO3
Mass of KClO3 = 49 g
Molar mass of KClO3 = 39 + 35.5 + 3(16) = 122.5 gmol−1
Molar mass of O2 = 2(16) = 32 gmol−1
Volume of O2 at STP = ?
No. of moles of KClO3 = mass/molar mass = 49/122.5 = 0.4 mol
Mole Comparison of KClO3 and O2
2KClO3 → 2KCl + 3O2
(Given) (Required)
2 mol --------------------> 3mol
0.4 mol ------------------> ?
According to unitary method
2 moles of KClO3 produces 3 mole O2
1 moles of KClO3 produces 3/2 mole O2
0.4 mole of KClO3 produces 3/2 x 0.4 mole O2 = 0.6 mol O2
Calculation of Volume O2 at STP
Volume of O2 at STP = n x molar volume = 0.6 x 22.4 = 13.44 dm3
Q3. How many grams and volume in cm3 of NH3 at STP are formed when 100 g of each of the following reagents are reacted together according to following equation;
2NH4Cl(s)+Ca(OH)2(s)→CaCl2(s)+2NH3(g)+ 2H2O
Solution
(I) Conversion of Mass of Each Reactant into Moles
Mass of NH4Cl = 100 g
Mass of Ca(OH)2 = 100 g
Molar mass of NH4Cl = 14 + 4 + 35.5 = 53.5 g/mol
Molar mass of Ca(OH)2= 40 + 32 + 2 = 74 g/mol
No. of moles (n) of NH4Cl = mass/molar mass = 100/53.5 = 1.869 mol
No. of moles (n) of Ca(OH)2 = mass/molar mass = 100/74 = 1.351mol
(II) Calculation of molar amount of product from molar amount of each reactant
2NH4Cl+ Ca(OH)2 →CaCl2 + 2NH3 + 2H2O
(Given) (Given) (Required)
2 mol 1 mol ----------------> 2 mole
1.869mol 1.315 mol ---------------> ?
Mole Comparison of NH4Cl and NH3
2 moles of NH4Cl produces 2 mole NH3
1 mole of NH4Cl produces 2/2 mole NH3
1.869 mole of NH4Cl produces 2/2 x 1.869 mole NH3 = 1.869 mole NH3
Mole Comparison of Ca(OH)2 and NH3
1 mole of Ca(OH)2 produces 2 mole NH3
1.351 mole of Ca(OH)2 produces 2/1 x 1.351 mole NH3 = 2.7 mole NH3
(
From above calculation, it is clear that least amount of product (NH3) is produced by NH4Cl so NH4Cl is a limiting reactant and amount of product is calculated from its amount.
(IV)Conversion of Molar Amount of Product into Gram
Molar amount of NH3= 1.869 moles
Molar mass of NH3 = 14 + 3(1) = 17 g/mol
Amount of NH3 in g = moles x molar mass =1.869 x 17 = 31.773 g NH3
(V) Conversion of Molar Amount of Product into Volume
Molar amount of NH3= 1.869 moles
Molar volume at STP in cm3 = 22400 cm3/mol
Volume of NH3 at STP = moles x molar volume =1.869 x 22400 = 41865.6 cm3 NH3
(iii) Volume in cm3 of 0.71 g Cl2 gas at STP
(iv) Number of formula units in 585 g of NaCl
Solution
Calculation of mass in gram of 5.6 dm3 of ethene (C2H4) at STP
Volume of ethene gas (C2H4) at STP = 5.6 dm3
Molar volume at STP = 22.4 dm3
No. of moles of given volume of ethene = ?
No. of moles (n) = Volume of gas at STP/Molar volume = 5.6/22.4 = 0.25 mol
Calculation of Number of molecules in 1 liter g of SO3
Volume of SO3 at STP = 1 liter
Molar volume at STP = 22.4 liter
Avogadro’s number = 6.02 x 1023 molecules/mol
No. of molecules of given volume of SO3 = ?
No. of molecules = V at STP/molar volume x NA = 1/22.4 x 6.02 x 1023 = 2.69 x 1022 molecules
Calculation of Volume in cm3 of 0.71 g Cl2 gas at STP
Mass of Cl2 gas at STP = 0.71 g
Molar mass of Cl2 gas = 2(35.5) = 71 g mol−1
Molar volume at STP = 22.4 dm3
No. of molecules = mass/molar mass x molar volume = 0.71/71 x 22.4 = 0.224 dm3
Calculation of Number of formula units in 585 g of NaCl
Mass of NaCl = 585 g
Molar mass of NaCl = 23 + 35.5 = 58.5 g mol−1
Avogadro’s number = 6.02 x 1023 formula units/mol
No. of molecules = mass/molar mass x NA = 585/58.5 x 6.02 x 1023 = 6.02 x 1024 formula units
Q2.Silver sulphide (Ag2S) is an anti microbial agent. In an experiment, 24.8g Ag2S is reacted with the excess of hydrochloric acid as given in the following reaction: (Atomic mass of Ag is 108 and S is 32)
Ag2S(s) + 2HCl(aq) → 2AgCl(s) + H2S(g)
Calculate the
(i)Mass of AgCl formed
(ii)Volume of H2S produced at STP
Solution
Calculation of moles of Ag2S
Mass of Ag2S = 24.8 g
Molar mass of Ag2S = 2(108) + 32 = 248 gmol−1
Molar mass of AgCl = 108 + 35.5 = 143.5 gmol−1
Molar volume at STP = 22.4 dm3
Mass of AgCl formed = ?
Volume of H2S at STP = ?
No. of moles of NH4NO3 = mass/molar mass = 24.8/248 = 0.1 mol
Mole Comparison of Ag2S and AgCl
Ag2S(s) + 2HCl(aq) → 2AgCl(s) + H2S(g)
(Given) (Required)
1 mol -----------------> 2mol
0.1 mol ---------------> ?
According to unitary method
1 moles of Ag2S produces 2 mole AgCl
0.1 mole of Ag2S produces 2 x 0.1 mole H2O = 0.2 mol AgCl
Calculation of Mass of AgCl
Mass = n x molar mass = 0.2 x 143.5 = 28.7 g AgCl
Calculation of Volume H2S at STP
Mole Comparison of Ag2S and H2S
Ag2S(s) + 2HCl(aq) → 2AgCl(s) + H2S(g)
(Given) (Required)
1 mol --------------------------> 1 mol
0.1 mol ------------------------> ?
According to unitary method
1 moles of Ag2S produces 1 mole H2S
0.1 mole of Ag2S produces 1 x 0.1 mole H2O = 0.1 mol H2S
Volume of H2S at STP = n x molar volume = 0.1 x 22.4 = 2.24 dm3 H2S
Q3. When aluminium is heated with nitrogen at 700oC, it gives aluminium nitride.
2Al(s) + N2(g) → 2AlN(s)
If 67.5 g of aluminium and 140 g of nitrogen gas are allowed to react, find out:
(a)Limiting reactant
(b) Mass of aluminum nitride produced?
(c) Mass of excess reactant?
Solution
(I) Conversion of Mass of Each Reactant into Moles
Mass of Al = 67.5 g
Mass of N2 = 140 g
Molar mass of Al = 27 g/mol
Molar mass of N2 = 2(14) = 28 g/mol
No. of moles (n) of NH4Cl = mass/molar mass = 67.5/27 = 2.5 mol
No. of moles (n) of Ca(OH)2 = mass/molar mass = 140/28 = 5.0 mol
(II) Calculation of molar amount of product from molar amount of each reactant
2Al(s) + N2(g) → 2AlN(s)
(Given) (Given) (Requreid)
2 mole 1 mole ----> 2 mole
2.5 mole 5.0 mole ---> ?
Mole Comparison of Al and AlN
2 moles of Al produces 2 mole AlN
1 mole of Al produces 2/2 mole AlN
2.5 mole of Al produces 2/2 x 2.5 mole AlN = 2.5 mole AlN
Mole Comparison of N2 and AlN
1 moles of N2 produces 2 mole AlN
5.0 mole of N2 produces 2 x 5.0 mole AlN = 10 mole AlN
(
From above calculation, it is clear that least amount of product (AlN) is produced by Al so Al is a limiting reactant and amount of product is calculated from its amount.
(IV) Conversion of Molar Amount of Product into Gram
Molar amount of AlN = 2.5 moles
Molar mass of AlN = 27 + 14 = 41 g/mol
Amount of AlN in g = moles x molar mass = 2.5 x 41 = 102.5 g AlN
(V) Calculation of Mass of Excess Reactant
2Al(s) + N2(g) → 2AlN(s)
(Given) (Requreid)
2 mole ------> 1 mole
2.5 mole -----> ?
Mole Comparison of Al and N2
2 moles of Al reacts with 1 mole N2
1 mole of Al produces ½ mole N2
2.5 mole of Al produces ½ x 2.5 mole N2 = 1.25 mole N2
Excess moles of N2 = Moles used – mole consumed = 5 – 1.25 = 3.75 moles N2 left behind
Conversion of excess mole into mass
Excess amount or mass of N2 = n x molar mass = 3.75 x 2(14) = 105 g N2 left behind
Q1. CaCO3 is often used to generate CO2 gas in industry. If 200 g of CaCO3 is strongly heated, what volume of CO2 gas will be obtained at 30°C and 1200 torr pressure? (KB- 2017)
Solution
Calculation of moles of CaCO3
Molar mass of CaCO3 = 40 + 12 + 48 = 100 gmol-1
Number of moles = mass/molar mass = 200/100 = 2 moles
Calculation of moles of CO2
CaCO3 → CaO + CO2
(Given) (Required)
1 mol ---------------------------> 1 mol
2 mol ----------------------------> ?
According to unitary method
1 mole of CaCO3 gives 1 mole of CO2
2 moles of CaCO3 gives = 1/1 x 2 = 2 moles of CO2
Calculation of Volume of CO2
n = 2 mole
T = 30°C ----> 30 + 273 = 303 K [ ∴K = °C + 273]
P = 1200 torr ----> 1200/760 = 1.58 atm [ ∴1 atm = 760 torr]
R = 0.0821 atm-dm3-mole–1-K–1
V = ?
PV = nRT
V= nRT/P
V = 2 (mole) x 0.0821 (atm-dm3-mol−1-K−1) x 303(K)/1.58 (atm)
V = 31.49 dm3
Q2. What volume of CO2 gas measured at 20°C and 720 torr pressure will be produced by the reaction between 200 g of Na2CO3 and HCl? (KB- 2016)
Na2CO3 + 2HCl → 2NaCl + CO2 + H2O
Solution
Calculation of moles of Na2CO3
Molar mass of Na2CO3 = 46 + 12 + 48 = 106 gmol−1
Number of moles = mass/molar mass = 200/106 = 1.886 moles
Calculation of moles of CO2
Na2CO3 + 2HCl → 2NaCl + CO2 + H2O
(Given) (Required)
1 mol ------------------------------> 1 mol
1.886 mol ----------------------> ?
According to unitary method
1 mole of Na2CO3 gives 1 mole of CO2
1.886 mole of Na2CO3 gives 1/1 x 1.886 = 1.886 mole of CO2
Calculation of Volume of CO2
n = 1.886 mole
T = 20°C ----> 20 + 273 = 293 K [ ∴K = °C + 273]
P = 720 torr ----> 720/760 = 0.947 atm [ ∴ 1 atm = 760 torr]
R = 0.0821 atm-dm3-mole–1-K–1
V = ?
PV = nRT
V= nRT/P
V = 1.886 (mole) x 0.0821 (atm-dm3-mol−1-K−1) x 293 (K)/0.947 (atm)
V = 47.90 dm3
Solution
Calculation of moles of KNO3
Molar mass of KNO3 = 39 + 14 + 48 = 101 gmol−1
Number of moles = mass/molar mass = 100/101 = 0.99 moles
Calculation of moles of O2
2KNO3 → 2KNO2 + O2
(Given) (Required)
2 mol ----------------> 1 mo
0.99 mol ------------> ?
According to unitary method
1 mole of KNO3 gives 1 mole of O2
0.99 mole of KNO3 gives ½ x 0.99 = 0.495 mole of O2
Calculation of Volume of O2
n = 0.495 mole
T = 39°C ----> 39 + 273 = 312 K [ ∴K = °C + 273]
P = 765 torr ----> 765/760 = 1.006 atm
[ ∴ 1 atm = 760 torr]
R = 0.0821 atm-dm3-mole–1-K–1
V = ?
PV = nRT
V= nRT/P
V = 0.495 (mole) x 0.0821 (atm-dm3-mol−1-K−1)
x 29303 (K)/1.006 (atm)
V = 12.06 dm3
Q4. If 53.5 g of NH4Cl is heated with Ca(OH)2, how many grams of NH3 is produced? Also find the volume of NH3 at STP according to the following equation (KB- 2021)
2NH4Cl + Ca(OH)2 →2NH3 + CaCl2 + 2H2O
Solution
Calculation of moles of NH4Cl
Molar mass of NH4Cl = 14 + (4x1) + 35.5 = 53.5 gmol−1
Number of moles = mass/molar mass = 53.5/53.5 = 1.00 moles
2NH4Cl + Ca(OH)2 → 2NH3 + CaCl2 + 2H2O
(Given) (Required)
2 mol -------------------> 2 mol
1 mol -------------------> ?
According to unitary method
2 mol of NH4Cl gives 2 mol of NH3
1 mol of NH4Cl gives = 2/2 x 1 = 1 mol of NH3
Calculation of Mass of NH3
Mass of NH3 = n x molar mass = 1 x 14+3 = 1 x 17 = 17 g
Calculation of Volume of NH3 at STP
Volume of NH3 at STP = n x molar volume = 1 x 22.4 dm3 = 22.4 dm3
Q5. Zinc reacts with H2SO4 (dil) as given below (KB- 2019, 2008)
Calculate the mass of ZnSO4, the volume of H2 gas at STP and the number of molecules of H2 gas which will be produced by reacting 163.5 g of Zn with H2SO4.
Solution
Calculation of moles of Zn
Molar mass of Zn = 65.4 gmol−1
Number of moles = mass/molar mass = 163.5/65.5 = 2.500 moles
Zn + H2SO4 → ZnSO4 + H2
Given Required
1 mole ---------------> 1 mole
2.5 mole -------------> ?
Calculation of moles of ZnSO4
According to unitary method
1 mol of Zn gives 1 mole of ZnSO4
2.5 mol of Zn gives = 2.5 x 1 mol of ZnSO4 = 2.5 mol of ZnSO4
Calculation of Mass of ZnSO4
Mass of ZnSO4 = n x molar mass = 2.5 x (65.4 + 32 + 16 x 4) = 1 x 17 = 161.4 g
Calculation of moles of H2
According to unitary method
1 mol of Zn gives 1 mole of H2
2.5 mol of Zn gives = 2.5 x 1 mol of H2 = 2.5 mol of H2
Calculation of Volume of H2 at STP
Volume of H2 at STP = n x molar volume = 2.5 x 22.4 = 56 dm3
Calculation of Number of molecules of H2
No. of molecules = n x Avogadro’s no. = 2.5 x 6.02 x 1023 = 1.505 x 1024 molecules
No comments:
Post a Comment