Carbon dioxide was passed through a solution weighing 250 g with a mass fraction of sodium hydroxide of 10 percent

Carbon dioxide was passed through a solution weighing 250 g with a mass fraction of sodium hydroxide of 10 percent. Determine the mass of carbon monoxide (4) that has reacted and the mass of the intermediate salt formed.

Given:
m solution (NaOH) = 250 g
ω (NaOH) = 10%

To find:
m (CO2) -?
m (salt) -?

Decision:
1) 2NaOH + CO2 => Na2CO3 + H2O;
2) M (NaOH) = 40 g / mol;
M (CO2) = 44 g / mol;
M (Na2CO3) = 106 g / mol;
3) m (NaOH) = ω * m solution / 100% = 10% * 250/100% = 25 g;
4) n (NaOH) = m / M = 25/40 = 0.63 mol;
5) n (CO2) = n (NaOH) / 2 = 0.63 / 2 = 0.32 mol;
6) m (CO2) = n * M = 0.32 * 44 = 14.08 g;
7) n (Na2CO3) = n (NaOH) / 2 = 0.63 / 2 = 0.32 mol;
8) m (Na2CO3) = n * M = 0.32 * 106 = 33.92 g.

Answer: The mass of CO2 is 14.08 g; Na2CO3 – 33.92 g.