Question

20 g of mixture of Na2Co3 and CaCo3 on heating produces 1.12 L of CO2 at stp. Find the % weight of Na2CO3 in the mixture

Answer 1

53.2% Na2CO3 and 46.8% NaHCO3

Explanation:

When you heat a mixture of Na2CO3 and NaHCO3, only the NaHCO3 decomposes.

2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)

The loss in mass is caused by the loss of H2O and CO2 (equivalent to the loss of H2CO3).

The loss in mass is (220 – 182) g = 38 g.

38g H2CO3 * 1 mol of H2CO3 / 62.02g of H2CO3 = 0.613 mol H2CO3

Moles of NaHCO3 = 0.613 mol H2CO3 * 2 mol NaHCO3 / 1 mol H2CO3 = 1.225 mol NaHCO3

Mass of NaHCO3 = 1.225 mol NaHCO3 * 84.01g NaHCO3 / 1 mol NaHCO3 = 102.9 g NaHCO3

The loss of mass after heating corresponds to 102.9 g of original NaHCO3.

The rest of the original mixture must have been Na2CO3.

Mass of original Na2CO3=220 g – 102.9 g=117.1 g

% of Na2CO3 = 117.1/ 220 * 100% = 53.2%

% of NaHCO3 = 102.9 / 220 * 100% = 46.8%