14.8 g of calcium hydroxide and 3.36 liters of carbon dioxide entered into the reaction. What is the mass of the precipitate formed?

univerkov | December 29, 2020 | education

Given:
m (Ca (OH) 2) = 14.8 g
V (CO2) = 3.36 l
Find: m (draft) -?
1) n (Ca (OH) 2) = m / M = 14.8 / 74 = 0.2 mol;
2) n (CO2) = V / Vm = 3.36 / 22.4 = 0.15 mol;
3) Ca (OH) 2 + CO2 => CaCO3 ↓ + H2O;
4) n (CaCO3) = n (CO2) = 0.15 mol;
5) m (CaCO3) = n * M = 0.15 * 100 = 15 g
Answer: The mass of CaCO3 is 15 g.

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