Phosphorus oxide (5), formed during the combustion of 3.1 g of phosphorus, solvents in 70 cm3

Phosphorus oxide (5), formed during the combustion of 3.1 g of phosphorus, solvents in 70 cm3, 14% solution of potassium hydroxide with a density of 1.14 g / cm3. Determine the composition of the obtained salt and its mass fraction in the resulting solution?

1.Let’s find the amount of phosphorus substance.

n = m: M.

M (P) = 31 g / mol.

n = 3.1 g: 31 g / mol = 0.1 mol.

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

4 P + 5 O2 = 2 P2O5.

According to the reaction equation, there are 2 mol of P2O5 per 4 mol of P. The substances are in quantitative ratios of 2: 1, the amount of substance P2O5 will be 2 times less than the amount of substance P.

n (P2O5) = ½ n (P) = 0.1: 2 = 0.05 mol.

Let’s find the mass of P2O5.

m = n × M.

M (P2O5) = 142 g / mol.

m = 142 g / mol × 0.05 mol = 7.1 g.

Find the mass of the KOH solution.

m = Vp.

m (KOH solution) = 70 cm3 × 1.14 g / cm3 = 79.8 g.

Let’s find the mass of KOH in the solution.

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

hence m (substance) = (m (solution) × w): 100%.

m (substance) = (79.8 g × 14%): 100% = 11.72 g.

Let’s find the amount of the substance KOH.

M (KOH) = 56 g / mol.

n = 11.72 g: 56 g / mol = 0.2 mol.

Let’s find the ratio of the amounts of substances KOH and P2O5.

n (KOH): n (P2O5) = 0.2: 0.05 = 4: 1 mol.

Let’s make an equation.

4KON + P2O5 = 2 K2NPO4 + H2O.

Salt – potassium hydrogen phosphate.

KOH and K2HPO4 are in quantitative ratios 4: 2 = 2: 1.

Let’s find the amount of substance К2НРО4.

M (K2HPO4) = 174 g / mol.

n (K2HPO4) = ½ n (KOH) = 0.2: 2 = 0.1 mol.

m (K2HPO4) = 174 g / mol × 01, mol = 17.4 g.

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

m (solution) = m (KOH solution) + m (P2O5).

m (solution) = 79.8 g + 7.1 g = 86.9 g.

W = (17.4g: 86.9g) x 100% = 20%.

Answer: 20%.



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