What volume of carbon dioxide will be released when nitric acid interacts with 50 g of limestone

What volume of carbon dioxide will be released when nitric acid interacts with 50 g of limestone containing 5% impurities?

Given:
m (limestone) = 50 g
ω approx. = 5%

To find:
V (CO2) -?

Decision:
1) 2HNO3 + CaCO3 => Ca (NO3) 2 + CO2 ↑ + H2O;
2) M (CaCO3) = Mr (CaCO3) = Ar (Ca) + Ar (C) + Ar (O) * 3 = 40 + 12 + 16 * 3 = 100 g / mol;
3) ω (CaCO3) = 100% – ω approx. = 100% – 5% = 95%;
4) m (CaCO3) = ω (CaCO3) * m (limestone) / 100% = 95% * 50/100% = 47.5 g;
5) n (CaCO3) = m (CaCO3) / M (CaCO3) = 47.5 / 100 = 0.475 mol;
6) n (CO2) = n (CaCO3) = 0.475 mol;
7) V (CO2) = n (CO2) * Vm = 0.475 * 22.4 = 10.64 l.

Answer: The volume of CO2 is 10.64 liters.