Question

A 12 g mixture of na2co3 and caco3 and clay when heated gave 1.792 l of co2 and in another experiment same amount of mixture required 19.8 l of 0.01 n hci for complete neutralization. Calculate weight of na2co3 present in sample.

Answer 1

Thus the % weight of Na2CO3 present in the sample is 33.3 %

Explanation:

Question statement:

A 12 g mixture of na2co3 and caco3 and clay when heated gave 1.792 l of co2 at STP. Calculate weight of na2co3 present in sample.?

Solution:

Reaction on heating

Na2CO3 ----> No effect

CaCO3 -----> CaO + CO2

CO2 is only produced by CaCO3 where there mole ratio are 1 : 1

At STP 1 mole gas = 22.4 litre volume; so

1.792 l at STP = 0.08 moles

So 0.08 moles of CO2 will be present in the sample.

Mass of CaCO3 = moles x molecular weight = 0.08 x 100 = 8 g

It means 4 g of NaNO3 will be present in the mixture.

Now % weight of Na2CO3 = wt. of compd in mixture / wt. of mixture x 100

% weight of Na2CO3 =  4 / 12 x 100

% weight of Na2CO3 =  0.333 x 100 = 33.3 %

Thus the % weight of Na2CO3 present in the sample is 33.3 %