CaCO3 + 2 HCl → CaCl2 + CO2 + H2O
10.0 g CaCO3 is .1 mole ... so .1 mole of CO2 should be produced
molar mass CO2 ... 44.0 g
% yield ... (3.65 / 4.40) * 100%
Marble (calcium carbonate) reacts with hydrochloric acid solution to form calcium
chloride solution, water and carbon dioxide. What is the percent yield of carbon dioxide
if 3.65 g of the gas is collected when 10.0 g marble reacts?
1 answer