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Question

What volume of CO2 at STP (atm) will be produced when 1 g of CaCO3 reacts with an excess of dilute HCl?

A
224 ml
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B
112 ml
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C
56 ml
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D
448 ml
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Solution

The correct option is A 224 ml
CaCO3+2HClCaCl2+CO2+H2O

1 mol of CaCO3 produces 1 mol of CO2=22.4 L of CO2 gas, i.e.
100 g of CaCO3 produces 22400 mL of CO2 gas.
1 g of CaCO3 will produce = 22400100 mL of CO2 gas.

Volume of CO2 gas produced = 224 mL

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