# 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%.

Answer: 14%; 6.86%.