Calculate the volume of ammonia that will be released when 890 g of ammonium nitrate, containing 20%

Calculate the volume of ammonia that will be released when 890 g of ammonium nitrate, containing 20% impurities, interacts with sodium hydroxide solution. How many grams of an 80% alkali solution will it take for the reaction?

This reaction proceeds according to the following chemical reaction equation:

NH4NO3 + NaOH = NH3 + NaNO3;

Find 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 = 890 x 0.8 / 80 = 8.9 mol;

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

Let’s define its volume.

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

V NH3 = 8.9 x 22.4 = 199.36 liters;

Calculate the weight of 8.9 mol of sodium hydroxide.

M NaOH = 23 + 16 +1 = 40 grams / mol;

m NaOH = 8.9 x 40 = 356 grams;

The weight of an 80% solution will be: 356 / 0.8 = 456 grams;