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Question

CaCO3+2HClCaCl2+H2O+CO2
The volume of CO2 gas formed when 2.5 g calcium carbonate are dissolved in excess hydrochloric acid at 0C and 1 atm pressure is :
(1 mole of any gas at 0C and 1 atm pressure occupies 22.414)

A
1.12 L
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B
56.0 L
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C
0.28 L
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D
0.56 L
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Solution

The correct option is C 0.56 L
No. of mole of calcium carbonate=given wt.molar mass=2.5100=0.025
1 mole of carbon dioxide gas has volume = 22.414 L
0.025 mol of carbon dioxide= 22.414 × 0.025= 0.56 L

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