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Question

What is the minimum mass of CaCO3(s) required to establish equilibrium in a 6.5 litres container for the following reaction (Kc=0.05 molL1)?

CaCO3(s)CaO(s)+CO2(g)

A
32.5 g
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B
24.6 g
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C
40.9 g
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D
8.0 g
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Solution

The correct option is A 32.5 g
Kc=[CO2]=0.05 mol/litre

So, moles of CO2=6.50×0.05 moles =0.3250 moles
CaCO3CaO+CO2

1 mole of CO2=1 mole of CaCO3

0.3250 moles of CO2=0.3250 moles of CaCO3=0.3250×100 g of CaCO3=32.5 g of CaCO3

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