A solution of 200 grams with a mass fraction of phosphoric acid of 9.8% Mixed with a solution of 200 g

A solution of 200 grams with a mass fraction of phosphoric acid of 9.8% Mixed with a solution of 200 g with a mass fraction of Na2HPO4 of 14.2% Determine the composition of the formed salt and its mass fraction in the solution

Determine the mass of the resulting salt using the following data:

Solution mass 1 – 200 g.

W – 9.8%.

Solution weight 2 – 200 g.

W 2 – 14.2%.

We write down the solution for this problem.

First, we write down the reaction equations.

Na2HPO4 + H3PO4 → 2 NaH2PO4.

Let us determine the mass and amount of the substance of the Na2HPO4 salt. We use the following formula.

m r.h. = W * m solution / 100%.

m = 200 * 0.098 = 19.6 g.

n = m / M.

M (Na2HPO4) = 23 * 2 + 1 + 31 + 16 * 4 = 142 g / mol.

n (Na2HPO4) = 19.6 / 142 = 0.14 mol.

m (H3PO4) = 200 * 0.142 = 28.4 g.

M (H3PO4) = 1 * 3 + 31 + 16 * 4 = 98 g / mol.

n = 28.4 / 98 = 0.29 mol.

Therefore, phosphoric acid is in excess.

0.14 mol Na2HPO4 – x mol NaH2PO4

1 mol – 2 mol.

X = 2 * 0.14: 1 = 0.28 mol.

m = 0.28 * 120 = 33.6 g.

W = 33.6 / (200 + 200) = 0.084 or 8.4%.

Answer: the mass fraction is equal to 8.4%.




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