100 g of a mixture of sodium carbonate and sodium bicarbanate was heated until the mass of the mixture no longer changed.

100 g of a mixture of sodium carbonate and sodium bicarbanate was heated until the mass of the mixture no longer changed. After heating, the mass of the residue became equal to 69 g. What is the percentage of the mixture?

Given:
m mixture (Na2CO3, NaHCO3) = 100 g
m (residue) = 69 g

To find:
ω (Na2CO3) -?
ω (NaHCO3) -?

Decision:
1) 2NaHCO3 = (tOC) => Na2CO3 + CO2 ↑ + H2O;
2) Let m1 (Na2CO3) = (x) r, then m1 (NaHCO3) = (100 – x) r;
3) n1 (NaHCO3) = m1 (NaHCO3) / M (NaHCO3) = ((100 – x) / 84) mol;
4) n2 (Na2CO3) = n1 (NaHCO3) / 2 = ((100 – x) / 84) / 2 = ((100 – x) / 168) mol;
5) m2 (Na2CO3) = n2 (Na2CO3) * M (Na2CO3) = ((100 – x) / 168) * 106 = ((100 – x) * 106/168) g;
6) m (residue) = m1 (Na2CO3) + m2 (Na2CO3);
69 = x + ((100 – x) * 106/168);
x = 16;
m1 (Na2CO3) = x = 16 g;
7) ω (Na2CO3) = m (Na2CO3) * 100% / m mixture (Na2CO3, NaHCO3) = 16 * 100% / 100 = 16%;
8) ω (NaHCO3) = 100% – ω (Na2CO3) = 100% – 16% = 84%.

Answer: Mass fraction of Na2CO3 is 16%; NaHCO3 – 84%.