To a solution weighing 200 g with a mass fraction of sulfuric acid of 8% was added a solution weighing 50 g

To a solution weighing 200 g with a mass fraction of sulfuric acid of 8% was added a solution weighing 50 g with a mass fraction of sodium hydroxide of 12%. What is the mass of sodium hydrogen sulfate that was isolated from the resulting solution?

Given:
m solution (H2SO4) = 200 g
ω (H2SO4) = 8%
m solution (NaOH) = 50 g
ω (NaOH) = 12%

To find:
m (NaHSO4) -?

1) H2SO4 + NaOH => NaHSO4 + H2O;
2) m (H2SO4) = ω (H2SO4) * m solution (H2SO4) / 100% = 8% * 200/100% = 16 g;
3) n (H2SO4) = m (H2SO4) / M (H2SO4) = 16/98 = 0.16 mol;
4) m (NaOH) = ω (NaOH) * m solution (NaOH) / 100% = 12% * 50/100% = 6 g;
5) n (NaOH) = m (NaOH) / M (NaOH) = 6/40 = 0.15 mol;
6) n (NaHSO4) = n (NaOH) = 0.15 mol;
7) m (NaHSO4) = n (NaHSO4) * M (NaHSO4) = 0.15 * 120 = 18 g.

Answer: The mass of NaHSO4 is 18 g.