Question

In the following reaction, what mass of magnesium carbonate, MgCO3, would be required to produce 989 L of carbon dioxide, CO2, measured at STP?

____________g?

Answer 1

Calculation of number of moles CO₂ produced :

Ideal gas law is PV=nRT

n = PV / RT

P = pressure = 1 atm

T = temperature = 273 K

R = gas constant = 0.0821 Latm/mol-K

V = volume = 989 L

Plug the values we get n = (1 x 989) / ( 0.0821 x 273)

                                   = 44.12 moles

The balanced equation is : MgCO₃ $\rightarrow$ MgO + CO₂

Frpom this balanced equation ,

1 mole of MgCO₃ produces 1 mole of CO₂

44.12 moles of MgCO₃ produces 44.12 moles of CO₂

Molar mass of MgCO₃ is = At.mass of Mg + At.mass of C + (3xAt.mass of O)

                                    = 24.3 + 12 + (3x16)

                                    = 84.3 g/mol

So mass of MgCO₃ is , m = number of moles x molarmass

                                      = 44.12 mol x 84.3 g/mol

                                      = 3719.3 g of MgCO₃

So mass of MgCO₃ required is 3719.3 g ~ 3719 g