Determine the mass of the reaction product, which can be obtained by the interaction

| September 20, 2021 | education

Determine the mass of the reaction product, which can be obtained by the interaction of 630 grams of nitric acid and 89.6 liters of ammonia.

Let’s find the amount of ammonia substance.

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

n = V: Vn.

n = 89.6 L: 22.4 L / mol = 4 mol.

Let’s find the amount of nitric acid substances.

M (HNO3) = 1 + 14 + 48 = 63 g / mol.

n = 630 g: 63 g / mol = 10 mol.

Nitric acid is given in excess. Solving the problem of deficiency (ammonia). Let’s compose the reaction equation. Let’s find the quantitative ratios of substances.

HNO3 + NH3 = NH4NO3.

1 mol of ammonia accounts for 1 mol of ammonium nitrate.

Substances are in quantitative ratios 1: 1.

The amount of substance will be equal.

n (NH3) = n (NH4NO3) = 4 mol.

Let’s find the mass of ammonium nitrate.

m = n × M.

M (NH4NO3) = 80 g / mol.

m = 80 g / mol × 4 mol = 320 g.

Answer: 320 g.



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