# The mass of sediment precipitated when passing carbon monoxide (4)

The mass of sediment precipitated when passing carbon monoxide (4) with a volume of 0.448 l through a solution containing ca (oh) 2 weighing 1.58 g is (g)

Ca (OH) 2 + CO2 = CaCO3 + H2O – ion exchange, precipitated calcium carbonate;
Calculations of the molar masses of substances:
M Ca (OH) 2 = 74 g / mol;

M (CO2) = 44 g / mol;

M (CaCO3) = 100 g / mol.

Let’s determine the amount of starting substances:
Y Ca (OH) 2 = m / M = 1.58 / 74 = 0.02 mol.

4. Proportion:

1 mole of gas at normal level – 22.4 liters;

X mol (CO2) – – 0.448 liters from here, X mol (CO2) = 1 * 0.448 / 22.4 = 0.02 mol.

The amount of starting materials is 0.02 mol.

Y (CaCO3) = 0.02 mol since all of these substances are 1 mol each.

We find the mass of the sediment:
m (CaCO3) = Y * M = 0.02 * 100 = 2 g

Answer: a precipitate of calcium carbonate weighing 2 g was obtained.

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