What volume of carbon monoxide (IV) is formed by the interaction of 60 g of calcium carbonate

What volume of carbon monoxide (IV) is formed by the interaction of 60 g of calcium carbonate containing 10% impurities with hydrochloric acid?

Reaction of calcium carbonate with hydrochloric acid.
CaCO3 + 2HCl = CaCl2 + CO2 + H2O.
Let’s determine the mass of impurities.
m (approx.) = 60 g • 10% / 100% = 6 g.
Mass of calcium carbonate without impurities.
m (CaCO3) = 60 g – 4 g = 54 g.
Determine the amount of moles of carbonate and carbon dioxide.
n (CaCO3) = n (CO2) = m / Mr = 54 g / 100 g / mol = 0.54 mol.
The gas volume is.
V (CO2) = n (CO2) • Vm = 0.54 mol • 22.4 mol / L = 12.1 L.



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