Calculate the mass of sulfuric acid required to neutralize a solution containing 10 g of sodium hydroxide.

2NaOh + H2SO4-> Na2SO4 + 2H2O
Given:
m (NaOh) = 10 g
Find:
m (H2SO4) -?

ν (nu) – amount of substance
ν = m / M

M (NaOh) = 23 + 16 + 1 = 40 g / mol
M (H2SO4) = 1 * 2 + 32 + 16 * 4 = 98 g / mol

According to the condition of the problem, 0.25 mol of sodium hydroxide NaOh enters into the reaction:
ν (NaOh) = 10g / 40g / mol = 0.25 mol

The reaction equation shows that 1 mol of sulfuric acid H2SO4 interacts with 2 moles of NaOh, that is, 2 times less than sodium hydroxide.
ν (H2SO4) = 0.25 mol / 2 = 0.125 mol

m = ν * M
m (H2SO4) = 0.125 mol * 98 g / mol = 12.25 g

Answer: To neutralize 10 grams of sulfuric acid H2SO4, 12.25 grams of sodium hydroxide NaOh is needed.