An excess of phosphoric acid solution was added to 296 g of a solution with a mass fraction

| February 13, 2021 | education

An excess of phosphoric acid solution was added to 296 g of a solution with a mass fraction of magnesium nitrate of 6%. Calculate the mass of the precipitate formed.

Reaction equation: 3Mg (NO3) 2 + 2H3PO4 = Mg3 (PO4) 2 + 6HNO3
The precipitate is magnesium orthophosphate.
Find the mass of the magnesium nitrate substance m in-va = w * m solution = 0.06 * 296 g = 17.76 g. Find the amount of magnesium nitrate substance n = m / M = 17.79 g / 148 g / mol = 0.12 mol. (M – molar mass, determined by the periodic table). According to the reaction, 1 mol of a precipitate is formed from 3 mol of magnesium nitrate, 0.12 / 3 = n (precipitate) / 1. n (sediment) = 0.12 / 3 = 0.04 mol. Find the mass of magnesium orthophosphate: m = n * M = 0.04 mol * 262 g / mol = 10.48 g.



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