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Question
Calcium carbonate react with dilute hydrochloric acid as given below:
\[\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}\]
What is the volume of carbon dioxide liberated at S.T.P. at the same time?
Solution
\[\ce{100 g of CaCO3 -> 22.4 l of CO2}\]
500 g of \[\ce{CaCO3}\] = `(22.4 xx 500)/100`
= 112.0
∴ 112.0 of \[\ce{CO2}\]
Hence, 112 l of \[\ce{CO2}\] is liberated from 5 moles of calcium carbonate.
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