# Calcium chloride solution was added to 150 g of sodium carbonate solution until precipitation ceased.

Calcium chloride solution was added to 150 g of sodium carbonate solution until precipitation ceased. The mass of the sediment was 12.0 g. Calculate the mass fraction of sodium carbonate in the original solution.

Given:
m solution (Na2CO3) = 150 g
m (sediment) = 12.0 g

Find:
ω (Na2CO3) -?

1) Na2CO3 + CaCl2 => CaCO3 ↓ + 2NaCl;
2) M (CaCO3) = Mr (CaCO3) = Ar (Ca) * N (Ca) + Ar (C) * N (C) + Ar (O) * N (O) = 40 * 1 + 12 * 1 + 16 * 3 = 100 g / mol;
3) n (CaCO3) = m / M = 12/100 = 0.12 mol;
4) n (Na2CO3) = n (CaCO3) = 0.12 mol;
5) M (Na2CO3) = Mr (Na2CO3) = Ar (Na) * N (Na) + Ar (C) * N (C) + Ar (O) * N (O) = 23 * 2 + 12 * 1 + 16 * 3 = 106 g / mol;
6) m (Na2CO3) = n * M = 0.12 * 106 = 12.72 g;
7) ω (Na2CO3) = m * 100% / m solution = 12.72 * 100% / 150 = 8.5%.

Answer: The mass fraction of Na2CO3 is 8.5%.

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