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OpenStudy (anonymous):
Calculate the mass of CaCO3 required to liberate 10 litres of CO2 at STP
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OpenStudy (anonymous):
CaCO3 → CaO + CO2 100 22.4 L 22.4 L of CO2 = 100 gm of CaCO3 10.0 L of CO2 = 100 x 10 / 22.4 = 44.64 gm of CaCO3
OpenStudy (anonymous):
ps because STP 22.4L=1MOLE
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