Calculate the mass of a solution of nitric acid with a mass fraction of 51 required

| April 29, 2021 | education

Calculate the mass of a solution of nitric acid with a mass fraction of 51 required to obtain copper (||) -nitate with a chemical amount of 0.1 mol from copper (||) -oxide.

Let’s write down given:

n (Cu (NO3) 2) = 0.1 mol

W (HNO3) = 51% or 0.51

m solution (HNO3) -?

Decision.

Let us write the equation for the reaction of the interaction of copper (II) oxide with nitric acid:

CuO + 2 HNO3 = Cu (NO3) 2 + H2O

According to the reaction equation, from 2 mol (HNO3) 1 mol (Cu (NO3) 2)

According to the condition of the problem, 0.1 mol (Cu (NO3) 2) is formed from x mol (HNO3)

X = 0.1 mol * 2 mol / 1 mol = 0.2 mol (HNO3)

Let’s calculate the mass of the formed acid:

M (HNO3) = 1 + 14 + 3 * 16 = 63 g / mol

m (HNO3) = 63 g / mol * 0.2 mol = 12.6 g

Calculate the mass of the acid solution:

W = m (substance) * 100% / m (solution)

m (solution) = m (substance) * 100% / W

m solution (HNO3) = 12.6 * 100% / 51% = 24.7 g

Answer: 24.7 g solution (HNO3)



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