Carbon monoxide (4) was passed through a solution of barium hydroxide with a mass of 200 g

Carbon monoxide (4) was passed through a solution of barium hydroxide with a mass of 200 g (mass fraction of hydroxide of 3.42%), barium bicarbonate was formed. Calculate the volume of gas that was passed through the solution.

Given:
m solution (Ba (OH) 2) = 200 g
ω (Ba (OH) 2) = 3.42%

Find:
V (CO2) -?

Solution:
1) Ba (OH) 2 + 2CO2 => Ba (HCO3) 2;
2) M (Ba (OH) 2) = Mr (Ba (OH) 2) = Ar (Ba) * N (Ba) + Ar (O) * N (O) + Ar (H) * N (H) = 137 * 1 + 16 * 2 + 1 * 2 = 171 g / mol;
3) m (Ba (OH) 2) = ω (Ba (OH) 2) * m solution (Ba (OH) 2) / 100% = 3.42% * 200/100% = 6.84 g;
4) n (Ba (OH) 2) = m (Ba (OH) 2) / M (Ba (OH) 2) = 6.84 / 171 = 0.04 mol;
5) n (CO2) = n (Ba (OH) 2) * 2 = 0.04 * 2 = 0.08 mol;
6) V (CO2) = n (CO2) * Vm = 0.08 * 22.4 = 1.792 l.

Answer: The CO2 volume is 1,792 liters.