What is the mass of the salt obtained by the interaction of a 300 g-10.2% solution

What is the mass of the salt obtained by the interaction of a 300 g-10.2% solution of sulfuric acid with sodium hydroxide?

Given:
m solution (H2SO4) = 300 g
ω (H2SO4) = 10.2%

Find:
m (salt) -?

1) H2SO4 + 2NaOH => Na2SO4 + 2H2O;
2) m (H2SO4) = ω * m solution / 100% = 10.2% * 300/100% = 30.6 g;
3) n (H2SO4) = m / M = 30.6 / 98 = 0.31 mol;
4) n (Na2SO4) = n (H2SO4) = 0.31 mol;
5) m (Na2SO4) = n * M = 0.31 * 142 = 44 g

Answer: The mass of Na2SO4 is 44 g.