An excess of calcium chloride solution was added to 848 g of a solution with a mass fraction of potassium

| May 1, 2021 | education

An excess of calcium chloride solution was added to 848 g of a solution with a mass fraction of potassium phosphate of 5%. Determine the mass of the precipitated sediment.

We draw up the reaction equation and place the coefficients: 2K3PO4 + 3CaCl2 = 6KCl + Ca3 (PO4) 2.
1) Find the mass of the potassium phosphate solute in its solution. To do this, we multiply the mass of the solution by the mass fraction of the substance and divide by one hundred percent: 848 * 5/100 = 42.4 g.
2) Find the amount of potassium phosphate substance. To do this, we divide the mass by the molar mass: 42.4 / 212 = 0.2 mol.
3) Find the amount of sediment substance. According to the reaction equation, there are two moles of potassium phosphate for one mole of calcium phosphate. This means that the amount of calcium phosphate substance is 0.1 mol.
4) Find the mass of the sediment: 0.1 * 310 = 31g – the answer.



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