Calculate the mass of salt that can be obtained from the reaction of magnesium (excess) with phosphoric acid weighing 24.5 g.

Let’s find the amount of phosphoric acid substance according to the formula: n = m: M.

M (H3PO4) = 3 + 31 + 64 = 98 g / mol.

n = 24.5 g: 98 g / mol = 0.25 mol.

Let’s compose the reaction equation, find in what quantitative ratios the substances are.

3MgO + 2H3PO4 = Mg3 (PO4) 2 + 3H2O

For 2 mol of phosphoric acid there is 1 mol of salt – magnesium phosphate. The substances are in quantitative ratios of 2: 1.

The amount of the salt substance will be 2 times less than that of the acid.

n (Mg3 (PO4) 2) = 2n (H3PO4) = 0.25: 2 = 0.125 mol.

Find the mass of salt by the formula m = n × M,

M (Mg3 (PO4) 2) = 24 × 3 = 2 × (31 + 4 × 16) = 262 g / mol.

m (Mg3 (PO4) 2) = 0.125 mol × 262 g / mol = 32.75 g.

Answer: m (Mg3 (PO4) 2) = 32.75 g.