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
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