What is the mass of salt obtained by the interaction of a 20% -foot solution of sulfuric acid weighing

What is the mass of salt obtained by the interaction of a 20% -foot solution of sulfuric acid weighing 300 g with sodium hydroxide weighing 80 g.

Given:
ω (H2SO4) = 20%
m solution (H2SO4) = 300 g
m (NaOH) = 80 g
Find: m (salt) -?
1) H2SO4 + 2NaOH => Na2SO4 + 2H2O;
2) m (H2SO4) = ω * m solution / 100% = 20% * 300/100% = 60 g;
3) n (H2SO4) = m / M = 60/98 = 0.6 mol;
4) n (NaOH) = m / M = 80/40 = 2 mol;
5) n (Na2SO4) = n (H2SO4) = 0.6 mol;
6) m (Na2SO4) = n * M = 0.6 * 142 = 85 g.
Answer: The mass of Na2SO4 is 85 g.