The action of excess hydrochloric acid on a sample of marble weighing 2.5 g

The action of excess hydrochloric acid on a sample of marble weighing 2.5 g released 448 ml of carbon dioxide. Mass fraction (%) of calcium carbonate in this sample was …

Given:
m mixture (CaCO3) = 2.5 g
V (CO2) = 448 ml = 0.448 l
To find:
w (CaCO3)
Decision:
CaCO3 + 2HCl = CaCl2 + H20 + CO2
n (CO2) = V / Vm = 0.448 l / 22.4 l / mol = 0.02 mol
n (CO2): n (CaCO3) = 1: 1
n (CaCO3) = 0.02 mol
m in islands (CaCO3) = n * M = 0.02 mol * 100 g / mol = 2 g
w (CaCO3) = m in / m mixture = 2 g / 2.5 g = 0.8 * 100% = 80%
Answer: 80%