Calculate the volume of carbon dioxide that will be released when 23 g of potassium carbonate

Calculate the volume of carbon dioxide that will be released when 23 g of potassium carbonate containing 10% impurities interacts with an excess of hydrochloric acid

Given:
m (K2CO3) = 23 g
ω approx. = 10%

To find:
V (CO2) -?

Decision:
1) K2CO3 + 2HCl => 2KCl + H2O + CO2 ↑;
2) ω (K2CO3) = 100% – ω approx. = 100% – 10% = 90%;
3) m clean. (K2CO3) = ω (K2CO3) * m (K2CO3) / 100% = 90% * 23/100% = 20.7 g;
4) n (K2CO3) = m (K2CO3) / M (K2CO3) = 20.7 / 138 = 0.15 mol;
5) n (CO2) = n (K2CO3) = 0.15 mol;
6) V (CO2) = n (CO2) * Vm = 0.15 * 22.4 = 3.36 l.

Answer: The CO2 volume is 3.36 liters.