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Q. 19 g of a mixture containing $ \text{NaHC}{{\text{O}}_{\text{3}}} $ and $ N{{a}_{2}}C{{O}_{3}} $ on complete heating liberated 1.12 L of $ C{{O}_{2}} $ at STP. The weight of the remaining solid was 15.9 g. What is the weight (in g) of $ N{{a}_{2}}C{{O}_{3}} $ in the mixture before heating?

Bihar CECEBihar CECE 2014Some Basic Concepts of Chemistry

Solution:

Molecular weight of $ NaHC{{O}_{3}}=23+1+12+48=84 $ Molecular weight of $ N{{a}_{2}}C{{O}_{3}}=46+12+48=106 $ Hence, total weight $ =84+106=190 $ $ \because $ In 190 g of a mixture, weight of $ NaC{{O}_{3}} $ is =106 $ \therefore $ In 19 g of a mixture weight of $ N{{a}_{2}}C{{O}_{3}}=\frac{106\times 19}{190} $ $ =10.6g $