In the laboratory, it was required to neutralize a solution containing 138 g of sulfuric acid

In the laboratory, it was required to neutralize a solution containing 138 g of sulfuric acid, how many grams of potassium hydroxide was required.

1. We write down the reaction of neutralization of sulfuric acid with potassium hydroxide:

H2SO4 + 2KOH = K2SO4 + H2O;

2.Calculate the chemical amount of sulfuric acid:

n (H2SO4) = m (H2SO4): M (H2SO4);

M (H2SO4) = 2 + 32 + 4 * 16 = 98 g / mol;

n (H2SO4) = 138: 98 = 1.4082 mol;

3. Determine the amount of potassium hydroxide and calculate its mass:

n (KOH) = n (H2SO4) * 2 = 1.4082 * 2 = 2.8164 mol;

m (KOH) = n (KOH) * M (KOH);

M (KOH) = 39 + 16 + 1 = 56 g / mol;

m (KOH) = 2.8164 * 56 = 157.72 g.

Answer: 157.72 g.