200 g of a 5% solution of hydrochloric acid reacted with a 25% solution of the same mass

200 g of a 5% solution of hydrochloric acid reacted with a 25% solution of the same mass of silver nitrate. Determine the mass fraction (%) of salt and acid remaining in the solution?

HCl + AgNO3 = AgCl + HNO3.

Mass of pure HCl in solution:

m (HCl) = 0.05 * 200 = 10 g.

Then the amount of acid:

n = 10 / 36.5 = 0.27397 mol.

Mass of pure AgNO3:

m (AgNO3) = 0.25 * 200 = 50 g.

Then the amount of silver nitrate:

n = 50/170 = 0.29411 mol.

Since silver nitrate is in excess (as can be seen from the above calculations), it will remain in solution after the reaction:

n (AgNO3) rest = 0.294 – 0.274 = 0.02 mol.

m (AgNO3) = 0.02 * 170 = 3.4 g.

Excess and deficiency problems are always solved due to deficiencies.

The amount of nitric acid is equivalent to the amount of hydrochloric acid, because it is in short supply:

n (HNO3) = 0.27397 mol.

m (HNO3) = 63 * 0.27397 = 17.26011 g.

m (solution) new will consist of the original solutions minus silver chloride.

m (solution) = 200 + 200- 0.27397 * 143.5 = 360.7 g.

w (HNO3) = 17.26011 / 360.68 = 4.8%.

w (AgNO3) = 3.4 / 360.7 = 9%.