Calculate the volume of ammonia that can be obtained by the interaction of ammonium

Calculate the volume of ammonia that can be obtained by the interaction of ammonium sulfate with a mass of 33 g and potassium hydroxide with a mass of 37 g.

The reaction of ammonium sulfate with potassium hydroxide is described by the following chemical reaction equation:

(NH4) 2SO4 + 2KOH = 2NH3 + K2SO4 + 2H2O;

One mole of ammonium sulfate reacts with two moles of potassium hydroxide. This produces two moles of ammonia.

Determine the amount of substance in 33 grams of ammonium sulfate:

M (NH4) 2SO4 = (14 + 4) x 2 + 32 + 16 x 4 = 132 grams / mol;

N (NH4) 2SO4 = 33/132 = 0.25 mol;

Determine the amount of substance in 37 grams of potassium hydroxide:

M KOH = 39 + 16 + 1 = 56 grams / mol;

N KOH = 37/56 = 0.61 mol;

In this reaction, 0.25 mol of ammonium sulfate is reacted with 0.25 x 2 = 0.5 mol of potassium hydroxide. This forms 0.25 x 2 = 0.5 mol of ammonia.

Let’s find its volume:

Under normal conditions, one mole of ideal gas takes up a volume of 22.4 liters.

V NH3 = 0.5 x 22.4 = 11.2 liters;