How many grams of phosphoric acid is required to react with 7.4g calcium hydroxide?

Let’s implement the solution:
In accordance with the condition of the problem, we write down the equation of the process:

3Ca (OH) 2 + 2H3PO4 = Ca3 (PO4) 2 + 6H2O – ion exchange, calcium phosphate precipitate was formed;

Calculations:
M Ca (OH) 2 = 74 g / mol;
M (H3PO4) = 97.9 g / mol.

Determine the amount of the original substance, if the mass is known:
Y Ca (OH) 2 = m / M = 7.4 / 74 = 0.1 mol.

Proportion:
0.1 mol Ca (OH) 2 – X mol (H3PO4);
-3 mol – 2 mol from here, X mol (H3PO4) = 0.1 * 2/3 = 0.06 mol.

We find the mass of acid:
m (H3PO4) = Y * M = 0.06 * 97.9 = 5.87 g

Answer: you need phosphoric acid weighing 5.87 g