Determine the volume of gas formed during the interaction of 0.2 kg and 0.1 l of 5%

Determine the volume of gas formed during the interaction of 0.2 kg and 0.1 l of 5% hydrochloric acid solution (its density is 0.9 g / cm ^ 3).

Given:
m (CaCO3) = 0.2 kg = 200 g
V solution (HCl) = 0.1 l = 100 ml
ω (HCl) = 5%
ρ solution (HCl) = 0.9 g / cm3 = 0.9 g / ml

To find:
V (gas) -?

1) CaCO3 + 2HCl => CaCl2 + CO2 ↑ + H2O;
2) n (CaCO3) = m / M = 200/100 = 2 mol;
3) m solution (HCl) = ρ solution * V solution = 0.9 * 100 = 90 g;
4) m (HCl) = ω * m solution / 100% = 5% * 90/100% = 4.5 g;
5) n (HCl) = m / M = 4.5 / 36.5 = 0.12 mol;
6) n (CO2) = n (HCl) / 2 = 0.12 / 2 = 0.06 mol;
7) V (CO2) = n * Vm = 0.06 * 22.4 = 1.3 liters.

Answer: The CO2 volume is 1.3 liters.