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Q.

Mass of CaCO3Volume of CO2 liberated
A) 200 g of CaCO31) 16.8 lit
B) 100 g of CaCO32) 11.2 lit
C) 50 g of CaCO33) 44.8 lit
D) 75 g of CaCO34) 22.4 lit

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a

A - 3, B - 4, C - 2, D - 1

b

A - 1, B - 2, C - 3, D - 4

c

A - 1, B - 3, C - 4, D - 2

d

A  - 3, B - 4, C - 1, D - 2

answer is B.

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Detailed Solution

The decomposition reaction of CaCO3 is given below.

 CaCO3(s)CaO (s) +   CO2(g)

1 mole of CaCO3  1 mole of CO2

Molar mass of CaCO3 =100 g

At STP,  100 g of CaCO3  22.4 lit of CO2

200g of CaCO3  22.4 L100 g×200 g = 44.8 lit of CO2

75g of CaCO3  22.4 L×75 g100 g=16.8 lit of CO2

50g of CaCO3  22.4100 g×50g=11.2 lit of CO2

            

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