To 300 g of a 4% solution of phosphoric acid was added 200 g of an 11% solution of silver nitrate

| June 11, 2021 | education

To 300 g of a 4% solution of phosphoric acid was added 200 g of an 11% solution of silver nitrate. Find the mass of the resulting sediment.

We compose the reaction equation and arrange the coefficients:
H3PO4 + 3AgNO3 = Ag3PO4 + 3HNO3.
The precipitate is silver phosphate.
1) The mass of the solute of orthophosphoric acid: 4 * 300/100 = 12g.
2) Amount of phosphoric acid substance: 12/98 = 0.12 mol.
3) The mass of the solute of silver nitrate: 11 * 200/100 = 22g.
4) The amount of silver nitrate solute: 22/170 = 0.13 mol.
5) Compare the amounts of phosphoric acid and silver nitrate substances. Lack of phosphoric acid. We carry out further calculations on it.
6) The amount of silver phosphate substance is 0.12 mol.
7) The mass of the sediment is 0.12 mol * 419 = 50.3 g – the answer.



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