The reaction involved 240 grams of calcium carbonate containing 5% impurities and some hydrochloric acid.

univerkov | December 9, 2020 | education

The reaction involved 240 grams of calcium carbonate containing 5% impurities and some hydrochloric acid. Find the volume of emitted carbon dioxide.

CaCO3 + 2HCl = CaCl2 + CO2 + H2O
find the mass of pure calcium carbonate
m (CaCO3) = 0.95 * 240 = 228 g
M (CaCO3) = 100 g / mol
Find the amount of calcium carbonate substance
n = m / M;
n = 228/100 = 2.28 mol
CaCO3: CO2 = 1: 1, then n (CO2) = 2.28 mol
V = n * Vm
Vm = 22.4 constant number for all gases
V = 2.28 * 22.4 = 51.1L

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