What mass of salt can be obtained by reacting 2.24 g of potassium hydroxide with nitric acid, if the yield is 96%?

1. The interaction of potassium hydroxide with nitric acid proceeds according to the equation:

KOH + HNO3 = KNO3 + H2O;

2.Calculate the chemical amount of potassium hydroxide:

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

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

n (KOH) = 2.24: 56 = 0.04 mol;

3.determine the theoretical amount of potassium nitrate:

ntheor (KNO3) = n (KOH) = 0.04 mol;

4. find the practical amount of salt obtained:

npract (KNO3) = ntheor (KNO3) * ν = 0.04 * 0.96 = 0.0384 mol;

5.calculate the mass of potassium nitrate:

m (KNO3) = npract (KNO3) * M (KNO3);

M (KNO3) = 39 + 14 + 3 * 16 = 101 g / mol;

m (KNO3) = 0.0384 * 101 = 3.8784 g.

Answer: 3.8784 g.