What volume of 10% (by weight) sulfuric acid (ml 1.07 g / ml) is required to neutralize a solution

What volume of 10% (by weight) sulfuric acid (ml 1.07 g / ml) is required to neutralize a solution containing 16.0 g of NaOH?

Given:
ω (H2SO4) = 10%
ρ solution (H2SO4) = 1.07 g / ml
m (NaOH) = 16.0 g

To find:
V solution (H2SO4) -?

1) H2SO4 + 2NaOH => Na2SO4 + 2H2O;
2) n (NaOH) = m (NaOH) / M (NaOH) = 16.0 / 40 = 0.4 mol;
3) n (H2SO4) = n (NaOH) / 2 = 0.4 / 2 = 0.2 mol;
4) m (H2SO4) = n (H2SO4) * M (H2SO4) = 0.2 * 98 = 19.6 g;
5) m solution (H2SO4) = m (H2SO4) * 100% / ω (H2SO4) = 19.6 * 100% / 10% = 196 g;
6) V solution (H2SO4) = m solution (H2SO4) / ρ solution (H2SO4) = 196 / 1.07 = 183.18 ml.

Answer: The volume of the H2SO4 solution is 183.18 ml.