A solution of silver nitrate was added to a solution weighing 730 g with a mass fraction

| April 5, 2021 | education

A solution of silver nitrate was added to a solution weighing 730 g with a mass fraction of hydrochloric acid of 10%. Calculate the mass of the precipitate formed.

As a result of the interaction of hydrochloric acid and silver nitrate, silver chloride (precipitate) and nitric acid are formed: HCl + AgNO3 = AgCl + HNO3.
1) Find the mass of the solute of hydrochloric acid in its solution. To do this, we multiply the mass of the solution by the mass fraction of the acid and divide by one hundred percent: 730 * 10/100 = 73g.
2) According to the periodic table, we find the molar mass of hydrochloric acid: 1 + 35.5 = 36.5.
3) The mass of the solute of hydrochloric acid: 73 / 36.5 = 2 mol.
4) According to the reaction equation, there is one mole of precipitate per mole of hydrochloric acid. This means that the amount of sediment substance is 2 mol.
5) Sludge weight: 2 * 143 = 286g – answer.



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