5.6 L of carbon dioxide was passed through the solution containing 14.8 g of calcium hydroxide.

5.6 L of carbon dioxide was passed through the solution containing 14.8 g of calcium hydroxide. determine the mass of the precipitate formed.

1. Let’s make the equation:
Ca (OH) 2 + CO2 = CaCO3 + H2O – compound reaction, precipitate of calcium carbonate is released;
2. Molar masses of substances:
M Ca (OH) 2 = 40 + 17 * 2 = 74 g / mol;
M (CaCO3) = 40 + 12 + 16 * 3 = 100 g / mol;
3. Let’s calculate the number of moles of calcium hydroxide, if the mass is known:
Y Ca (OH) 2 = m / M = 14.8 / 74 = 0.2 mol;
4. Let’s calculate the amount of moles of carbon dioxide:
1 mol of gas at n. y – 22.4 liters.
X mol (CO2) -5.6 l. hence, X mol (CO2) = 1 * 5.6 / 22.4 = 0.25 mol;
5. Let’s make a proportion, taking into account that the calculation is carried out for the substance in deficiency:
0.2 mol of Ca (OH) 2 – X mol (CaCO3);
-1 mol                           -1 mol hence, X mol (CaCO3) = 0.2 * 100 = 20 g.
Answer: during the reaction, calcium carbonate weighing 20 g is released.