23.17 liters of ammonia under normal conditions were passed through 160.17 milliliters of a 30%

| June 19, 2021 | education

23.17 liters of ammonia under normal conditions were passed through 160.17 milliliters of a 30% nitric acid solution with a density of 1.18 grams / milliliter. Determine the mass of the formed salt.

1. Let’s write down the reaction equation:

NH3 + HNO3 = NH4NO3.

2. Find the amount of ammonia, mass and amount of nitric acid:

n (NH3) = V (NH3) / Vm = 23.17 L / 22.4 L / mol = 1.03 mol.

m (solution HNO3) = V (solution HNO3) * ρ (solution HNO3) = 160.17 ml * 1.18 g / ml = 189 g.

m (HNO3) = (m (solution) * ω (HNO3)) / 100% = (189 g * 30%) / 100% = 56.7 g.

n (HNO3) = m (HNO3) / M (HNO3) = 56.7 g / 63 g / mol = 0.9 mol.

3. Since according to the equation we have nitric acid in short supply, we will look for salt by the amount of this acid:

n (NH4NO3) = n (HNO3) = 0.9 mol.

m (NH4NO3) = n (NH4NO3) * M (NH4NO3) = 0.9 mol * 80 g / mol = 72 g.

4. Let us find the mass of the reacted ammonia, the mass of the solution and the mass fraction of the salt.

nreacted (NH3) = n (HNO3) = 0.9 mol.

m (NH3) = n (NH3) * M (NH3) = 0.9 mol * 17 g / mol = 15.3 g.

m (solution) = m (NH3) + m (solution HNO3) = 15.3 g + 189 g = 204.3 g.

ω (NH4NO3) = (m (NH4NO3) / m (solution)) * 100% = 72 g / 204.3 g * 100% = 35.24%.

Answer: 35.24%.



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