Question

Dr. Dahm is trying to generate CO_2(g) by reacting an acid with calcium carbonate. If he pours 389 mL of a 1.6 M solution of H₂SO₄ into a flask that contains 69.5 g of CaCO₃. If the temperature is 30 °C and the pressure is 0.984 atm calculate the volume of CO_2(g) produced.

                        H₂SO₄    +    CaCO₃ CO_2(g)   +      CaSO_4(aq)    +    H₂O_(l)

Answer 1

The balanced equation is :   H₂SO₄ + CaCO₃ CO_2(g)   + CaSO_4(aq) + H₂O_(l)

Number of moles of H₂SO₄ , n = MOlarity x volume in L

                                                  = 1.6 M x ( 389 / 1000 ) L

                                                 = 0.6224 mol

Molar mass of CaCO₃ is = At.mass of Ca + At.mass of C + (3xAt.mass of O )

                                        = 40 + 12+(3x16)

                                        = 100 g/mol

Given mass of CaCO₃ is , m = 69.5 g

So number of moles of CaCO₃ is ,n' = mass/molar mass

                                                           = 69.5 g / 100 (g/mol)

                                                           = 0.695 mol

According to the balanced equation ,

1 mole of H₂SO₄ reacts with 1 mole of CaCO₃

0.6224 mole of H₂SO₄ reacts with 0.6224 moles of CaCO₃

So 0.695 - 0.6224 = 0.0726 moles of CaCO₃ left unreacted in the solution

Since all the mass of H₂SO₄ completly reacted it is the limiting reactant

From the balanced equation ,

1 mole of CaCO₃ produces 1 mole of CO₂

0.6224 mole of CaCO₃ produces 0.6224 mole of CO₂

We know that PV = nRT

Where

T = Temperature = 30 oC = 30+273 = 303 K

P = pressure = 0.984 atm

n = No . of moles = 0.6224 mol

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the gas = ?

Plug the values we get V = nRT / P

                                         = ( 0.6224 x 0.0821 x 303 ) / 0.984

                                        = 15.7 L

Therefore the volume of CO₂ produced is 15.7 L