4.4 g of CO2 were passed into lime water (Ca (OH) 2). What is the mass of the resulting sludge?

Let’s write the reaction equation:

Ca (OH) 2 + CO2 = CaCO3 ↓ + H2O

Let’s find the amount of substance carbon dioxide:

v (CO2) = m (CO2) / M (CO2) = 4.4 / 44 = 0.1 (mol).

According to the reaction equation, from 1 mol of CO2, 1 mol of CaCO3 is formed, therefore:

v (CaCO3) = v (CO2) = 0.1 (mol).

Thus, the mass of the resulting precipitate of calcium carbonate:

m (CaCO3) = v (CaCO3) * M (CaCO3) = 0.1 * 100 = 10 (g).

Answer: 10 g.