Find the mass of sulfuric acid required to neutralize 250 g of 15% lithium hydroxide.

Let us find the mass of LiOH in solution.

W = m (substance): m (solution) × 100%,

m (substance) = (m (solution) × W): 100%.

m (LiOH) = (250 g × 15%): 100% = 37.5 g.

Let’s find the amount of the substance LiOH.

M (LiOH) = 24 g / mol.

n = m: M.

n = 37.5 g: 24 g / mol = 1.56 mol.

Let’s compose the reaction equation. Let’s find the quantitative ratios of substances.

2LiOH + H2SO4 = Li2SO4 + 2H2O.

According to the reaction equation, there is 1 mol of H2SO4 for 2 mol of LiOH. Substances are in quantitative ratios of 2: 1.

The amount of H2SO4 is 2 times less than the amount of LiOH.

n (H2SO4) = ½ n (LiOH) = 1.56: 2 = 0.78 mol.

Let’s find the mass of H2SO4.

m = n × M,

M (H2SO4) = 98 g / mol.

m = 0.78 mol × 98 g / mol = 76.44 g.

Answer: 76.44 g.