0.336 L of carbon dioxide was passed through the solution containing 1.48 g of calcium hydroxide.

0.336 L of carbon dioxide was passed through the solution containing 1.48 g of calcium hydroxide. Find the mass of the sediment.

Given:
m (Ca (OH) 2) = 1.48 g
V (CO2) = 0.336 l

To find:
m (draft) -?

Decision:
1) Ca (OH) 2 + CO2 => CaCO3 ↓ + H2O;
2) M (Ca (OH) 2) = Mr (Ca (OH) 2) = Ar (Ca) * N (Ca) + Ar (O) * N (O) + Ar (H) * N (H) = 40 * 1 + 16 * 2 + 1 * 2 = 74 g / mol;
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 (Ca (OH) 2) = m (Ca (OH) 2) / M (Ca (OH) 2) = 1.48 / 74 = 0.02 mol;
4) n (CO2) = V (CO2) / Vm = 0.336 / 22.4 = 0.015 mol;
5) n (CaCO3) = n (CO2) = 0.015 mol;
6) m (CaCO3) = n (CaCO3) * M (CaCO3) = 0.015 * 100 = 1.5 g.

Answer: The mass of CaCO3 is 1.5 g.



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