# By passing 5.6 liters of ammonia through a solution of sulfuric acid weighing 98 g

**By passing 5.6 liters of ammonia through a solution of sulfuric acid weighing 98 g with a mass fraction of 7%, ammonium sulfate was formed. Calculate the mass of the resulting salt.**

Ammonia reacts with sulfuric acid. This synthesizes the ammonium sulfate salt.

This reaction proceeds according to the following chemical reaction equation:

2NH3 + H2SO4 = (NH4) 2SO4;

Let’s calculate the chemical amount of ammonia. To do this, we divide its volume by the standard volume of 1 mole of gas (which is 22.40 liters)

N NH3 = 5.6 / 22.4 = 0.25 mol;

Let’s find the chemical amount of sulfuric acid. For this purpose, we divide its weight by its molar mass.

M H2SO4 = 2 + 32 + 16 x 4 = 98 grams / mol;

N H2SO4 = 98 x 0.93 / 98 = 0.93 mol;

Sulfuric acid is taken in excess.

0.25 mol of ammonia will react with 0.125 mol of sulfuric acid. This will synthesize 0.125 mol of ammonium sulfate.

Let’s determine its weight.

To do this, we multiply the chemical (molar) amount of a substance by its molar (atomic) mass.

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

The weight of ammonium sulfate will be:

m (NH4) 2SO4 = 132 x 0.125 = 16.5 grams;