Calculate the mass of a solution with a mass fraction of nitric acid of 20% required

| September 21, 2021 | education

Calculate the mass of a solution with a mass fraction of nitric acid of 20% required to dissolve 408 g of aluminum oxide with a mass fraction of impurities of 25%.

1. Let’s compose the equation of a chemical reaction.

6HNO3 + Al2O3 = 2Al (NO3) 3 + 2H2O.

2. Find the amount of aluminum oxide.

ω (Al2O3) = 100% – ω (impurities) = 100% – 25% = 75%.

m (Al2O3) = m (mixtures) * ω (Al2O3) / 100% = 408 g * 75% / 100% = 306 g.

n (Al2O3) = m (Al2O3) / M (Al2O3) = 306 g / 102 g / mol = 3 mol.

3. Using the reaction equation, we find the amount of nitric acid and, as a result, the mass of the required solution.

n (HNO3) = n (Al2O3) * 6 = 3 mol * 6 = 18 mol.

m (HNO3) = n (HNO3) * M (HNO3) = 18 mol * 63 g / mol = 1134 g.

m (solution) = m (HNO3) * 100% / ω (HNO3) = 1134 g * 100% / 20% = 56700 g.

Answer: m (solution) = 56700 g.



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