Calculate the mass of the salt if 50 g of potassium hydroxide has reacted with 100 g of sulfuric acid.

Let’s write down given:

m (KOH) = 50 g

m (H2SO4) = 100 g

m (salt) -?

Solution:

1) Calculate the amount of substance of each of the reacting compounds:

n = m / M;

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

n (KOH) = 50 g / 56 g / mol

n (KOH) = 0.89 mol

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

n (H2SO4) = 100 g / 98 g / mol

n (H2SO4) = 1.02 mol

2) Let us write down the equation of the ongoing reaction with the formation of an average salt and determine whether KOH is enough for this reaction:

2 KOH + H2SO4 = K2SO4 + 2 H2O

According to the reaction, 1 mol of acid requires 2 mol of alkali;

Therefore, 1.02 mol of acid requires 2.08 mol of alkali, we have only 0.89 mol of alkali. Alkali is deficient and acidic salt will form as a result.

3) We write down the equation of the ongoing reaction with the formation of an acidic salt and calculate its mass:

KOH + H2SO4 = KHSO4 + H2O

M (KHSO4) = 136 g / mol

56 g (KOH) -136 g (KHSO4)

50 g (KOH) – x g (KHSO4)

X = 50g * 136g / 56g

X = 121.4 g (KHSO4)

Answer: 121.4 g (KHSO4)