Question

Part A: When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction CaCO3(s)→CaO(s)+CO2(g) What is the mass of calcium carbonate needed to produce 85.0 L of carbon dioxide at STP?

Part B: Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is 2C4H10(g)+13O2(g)→8CO2(g)+10H2O(l) At 1.00 atm and 23 ∘C, what is the volume of carbon dioxide formed by the combustion of 1.80 g of butane?

Answer 1

Ans :

Part A :

Each mol of gas occupies 22.4 L volume at STP

so the number of mol of gas occupying 85 L of volume will be :

= 85 / 22.4 = 3.79 mol

Each mol CO2 is produced from 1 mol CaCO3

So number of mol of CaCO3 required = 3.79 mol

So mass of CaCO3 = mol x molar mass

= 3.79 mol x 100.0869 g/mol

= 379.3 grams

Part B :

Number of mol of butane = mass / molar mass

= 1.80 g / 58.12 g/mol = 0.031 mol

Each mol butane makes 4 mol CO2

So number of mol of CO2 formed will be : 0.031 mol x 4

= 0.124 mol

using ideal gas law pV = nRT

p = pressure = 1.00 atm

n = 0.124 mol

R = ideal gas constant = 0.0821 L.atm/mol.K

T = 23 C = 296.15 K

putting all values :

1.00 atm x V = 0.124 mol x 0.0821 L.atm/mol.K x 296.15 K

V = 3.01 L

So the volume of carbon dioxide formed will be 3.01 L