What volume of ammonia will be released during the interaction of 830 grams of ammonium

What volume of ammonia will be released during the interaction of 830 grams of ammonium nitrate containing 20% impurities with sodium hydroxide solution.

This reaction proceeds according to the following chemical reaction equation:

NH4NO3 + NaOH = NH3 + NaNO3;

Let’s calculate the chemical amount of ammonium nitrate. For this purpose, we divide its weight by its molar mass.

M NH4NO3 = 14 + 4 + 14 + 16 x 3 = 80 grams / mol;

N NH4NO3 = 830 x 0.8 / 80 = 8.3 mol;

During the reaction, the same volume of ammonia will be released.

Let’s calculate its volume.

To do this, multiply the amount of substance by the standard volume of 1 mole of gas (filling 22.40 liters)

V NH3 = 8.3 x 22.4 = 185.92 liters;