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When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction CaCO3(s)→CaO(s)+CO2(
g. What is the mass of calcium carbonate needed to produce 27.0 L of carbon dioxide at STP?

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Dejanras
Answer is: mass of calcium carbonate needed is 120 grams.
Chemical reaction: CaCO₃(s) → CaO(s) + CO₂(g).
V(CO₂) = 27.0 L.
Vm = 22.4 L/mol.
n(CO₂) = V(CO₂) ÷ Vm.
n(CO₂) = 27 L ÷ 22.4 L/mol.
n(CO₂) = 1.2 mol.
From chemical reaction: n(CO₂) : n(CaCO₃) = 1 : 1.
m(CaCO₃) = 1.2 mol.
m(CaCO₃) = n(CaCO₃) · M(CaCO₃).
m(CaCO₃) = 1.2 mol · 100 g/mol.
m(CaCO₃) = 120 g.

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