Through 200 g of a solution with a mass fraction of acetic acid of 30%, 6, 72 liters of ammonia were passed.

Through 200 g of a solution with a mass fraction of acetic acid of 30%, 6, 72 liters of ammonia were passed. Calculate the mass of sodium hydroxide that can react in an acid solution after passing black ammonia.

Given:
m solution (CH3COOH) = 200 g
ω (CH3COOH) = 30%
V (NH3) = 6.72 L

To find:
m (NaOH) -?

Decision:
1) CH3COOH + NH3 => CH3COONH4;
CH3COOH + NaOH => CH3COONa + H2O;
2) m (CH3COOH) = ω (CH3COOH) * m solution (CH3COOH) / 100% = 30% * 200/100% = 60 g;
3) n total (CH3COOH) = m (CH3COOH) / M (CH3COOH) = 60/60 = 1 mol;
4) n (NH3) = V (NH3) / Vm = 6.72 / 22.4 = 0.3 mol;
5) n1 (CH3COOH) = n (NH3) = 0.3 mol;
6) n2 (CH3COOH) = n total. (CH3COOH) – n1 (CH3COOH) = 1-0.3 = 0.7 mol;
7) n (NaOH) = n2 (CH3COOH) = 0.7 mol;
8) m (NaOH) = n (NaOH) * M (NaOH) = 0.7 * 40 = 28 g.

Answer: The mass of NaOH is 28 g.