Stoichiometric Calculation (Numericals solution)

 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)

             CaCO₃      →    CaO         +      CO₂

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

Q3. 100 grams of KNO3 are heated to redness. What volume of oxygen at 39°C and 765 mm pressure will evolve? 2KNO3 → 2KNO2 + O2

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)

Zn        +  H₂SO₄     →    ZnSO₄     +  H₂

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