How many grams of barium carbonate is formed by passing 5.6 liters of carbon dioxide

How many grams of barium carbonate is formed by passing 5.6 liters of carbon dioxide through a barium hydroxide solution?

Given:
V (CO2) = 5.6 l

Find:
m (BaCO3) -?

Solution:
1) Write the reaction equation:
CO2 + Ba (OH) 2 => BaCO3 ↓ + H2O;
2) Calculate the amount of substance CO2:
n (CO2) = V (CO2) / Vm = 5.6 / 22.4 = 0.25 mol;
3) Determine the amount of substance BaCO3:
n (BaCO3) = n (CO2) = 0.25 mol;
4) Calculate the molar mass of BaCO3:
M (BaCO3) = Mr (BaCO3) = Ar (Ba) * N (Ba) + Ar (C) * N (C) + Ar (O) * N (O) = 137 * 1 + 12 * 1 + 16 * 3 = 197 g / mol;
5) Calculate the mass of BaCO3:
m (BaCO3) = n (BaCO3) * M (BaCO3) = 0.25 * 197 = 49.25 g.

Answer: The mass of BaCO3 is 49.25 g.