# To 500 g of a solution containing sulfuric and nitric acids, an excess of barium chloride solution was added.

To 500 g of a solution containing sulfuric and nitric acids, an excess of barium chloride solution was added. A precipitate was separated, the mass of which was 81.55 g, the resulting solution was neutralized. For neutralization, 140.8 ml of sodium hydroxide solution with a mass fraction of 25% and a density of 1.25 g / ml were spent. Write the reaction equations. Determine the mass of sulfuric acid. Determine the mass of nitric acid. Determine the mass fractions (in percent) of both acids in the solution.

Let’s find the amount of BaSO4 sediment substance according to the formula:

n = m: M.

M (BaSO4) = 233 g / mol.

n = 81.55 g: 233 g / mol = 0.35 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

H2SO4 + BaCl2 = BaSO4 + 2HCl.

According to the reaction equation, there is 1 mol of H2SO4 for 1 mol of BaSO4. The substances are in quantitative ratios of 1: 1.

n (H2SO4) = n (BaSO4) = 0.35 mol.

M (H2SO4) = 98 g / mol.

m = 0.35 mol × 98 g / mol = 34.3 g.

n (HCl) = n (BaSO4) = 0.35 mol.

HCl + NaOH = NaCl + H2O.

HNO3 + NaOH = NaNO3 + H2O.

Find the mass of the NaOH solution.

m = Vp.

m = 140.8 ml × 1.25 g / ml = 176 g.

W = m (substance): m (solution) × 100%,

m (substance) = (m (solution) × W): 100%.

m (NaOH) = (176 g × 25%): 100% = 44 g.

M (NaOH) = 40 g / mol.

n = 44 g: 40 g / mol = 1.1 mol.

In the reaction with HCl, 0.7 mol of NaOH reacts (n (HCl) = n (NaOH) = 0.7 mol.

0.4 mol of the remaining NaOH is reacted with HNO3.

According to the reaction equation, 1 mol of NaOH accounts for 1 mol of HNO3. The substances are in quantitative ratios of 1: 1.

n (HNO3) = n (NaOH) = 0.4 mol.

M (HNO3) = 63g / mol.

m = 0.4 mol × 63 g / mol = 25.2 g.

W (HNO3) = (25.2: 500 g) × 100% = 5.04%.

W (H2SO4) = (34.3: 500 g) × 100% = 6.86%. 