What salt and what mass is formed when 4.48 liters of ammonia is passed through 196 grams

What salt and what mass is formed when 4.48 liters of ammonia is passed through 196 grams of 10% phosphoric acid solution?

V (NH3) = 4.48 l.

m (H3PO4) = 196 g.

W = 10%.

Let’s determine the mass of the formed salt. We write down the solution.

First, we find the amount of ammonia and phosphoric acid.

n (NH3) = V / Vm.

n (NH3) = 4.48 / 22.4 = 0.2 mol.

m (H3PO4) = 196 * 0.1 = 19.6 g.

n = m / M.

M (H3PO4) = 98 g / mol.

n (H3PO4) = 19.6 / 98 = 0.2 mol.

This means that the following salt is formed as a result, we write down the reaction equation.

NH3 + H3PO4 = NH4H2PO4

We find the mass of the formed salt.

0.2 mol – x mol salt

1 mol – 1 mol.

X = 1 * 0.2: 1 = 0.2 mol.

m = n * M.

M (NH4H2PO4) = 115 g / mol.

m (NH4H2PO4) = 0.2 * 115 = 23 g.

The result is a salt called ammonium dihydrogen phosphate.

Answer: ammonium dihydrogen phosphate, weighing 23 grams.