How much ammonia will be released when 8.5 g of ammonium nitrate interacts

How much ammonia will be released when 8.5 g of ammonium nitrate interacts with sodium hydroxide solution? How many grams of a 30% alkali solution is required for this reaction?

The reaction of ammonium nitrate with sodium hydroxide is described by the following chemical reaction equation:

NH4NO3 + NaOH = NH3 + NaNO3 + H2O;

One mole of ammonium nitrate reacts with one mole of sodium hydroxide. This produces one mole of ammonia.

Determine the amount of substance in 8.5 grams of ammonium nitrate:

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

N NH4Cl = 8.5 / 80 = 0.10625 mol;

Therefore, during the reaction, 0.10625 mol of ammonia was formed.

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

N NH3 = 0.10625 x 22.4 = 2.38 liters;

Find the required mass of 30% sodium hydroxide solution.

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

m NaOH = 40 x 0.10625 = 4.25 grams;

The required mass of the solution will be:

M solution = 4.25 / 0.30 = 14.17 grams;