Determine the mass of the precipitate formed by the interaction of 980 g of a 25% phosphoric acid

| March 17, 2021 | education

Determine the mass of the precipitate formed by the interaction of 980 g of a 25% phosphoric acid solution with calcium hydroxide.

Let’s find the mass of Н3РО4 in solution.

W = m (substance): m (solution) × 100%,

m (substance) = (m (solution) × W): 100%.

m (H3PO4) = (980 g × 25%): 100% = 245 g.

Let’s find the amount of substance Н3РО4.

M (H3PO4) = 98 g / mol.

n = m: M.

n (H3PO4) = 245 g: 98 g / mol = 2.5 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

2H3PO4 + 3Ca (OH) 2 = Ca3 (PO4) 2 ↓ + 6H2O.

According to the reaction equation, there is 1 mol of Ca3 (PO4) 2 per 2 mol of Н3РО4. The substances are in quantitative ratios of 2: 1. The amount of Ca3 (PO4) 2 is 2 times less than the amount of H3PO4.

n (Ca3 (PO4) 2) = ½ n (H3PO4) = 2.5: 2 = 1.25 mol.

Let’s find the mass of Ca3 (PO4) 2.

m = n × M,

M (Ca3 (PO4) 2) = 310 g / mol.

m (Ca3 (PO4) 2) = 310 g / mol × 1.25 mol = 387.5 g.

Answer: 387.5 g.



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