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  • 20 g mixture of Na2CO3 and CaCO3 is gently

20 g mixture of Na2CO3 and CaCO3 is gently heated and 2.24 litre of CO2 at STP is produced. What is the % weight of Na2CO3 in that sample?

Option: 1

12.5%


Option: 2

25%


Option: 3

50%


Option: 4

75%


Answers (1)

best_answer

We know that Na2CO3 does not decompose on heating. So CO2 will be produced by only CaCO3

CaCO3(s)→CaO(s) + CO2(g)

By reaction stoichiometry

Mole of CaCO3 = mole of CO2

Mole of CaCO3 = 2.24 litre

Mole of CaCO3 =  2.24/22.4 mole

Mole of CaCO3 =  0.1 mole

Now,

Weight of CaCO3 =  mole X molar mass of CaCO3  

Weight of CaCO3 =  0.1 X 100 = 10g

wt. of Na2CO3 =  20g - wt. of CaCO3

wt. of Na2CO3 =  20g - 10 = 10g

% wt. of Na2CO3 =  (10/20)X100 % =  50%

Therefore, Option(3) is correct.

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