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%.