Question

Calculate the mass of H3PO4 that will be needed to produce
480 grams of water in the following reaction?

__Ca(OH)2 + __H3(PO4) → Ca3(PO4)2 + __H2O

Answer 1

1) Balance the equation:
3 Ca(OH)2 + 2 H3PO4 → Ca3(PO4)2 + 6 H2O

2) Stoichiometry:
2 mol H3PO4 → 6 mol H2O, so 1 mol H3PO4 → 3 mol H2O.
Moles H3PO4 needed = moles H2O / 3.

3) Moles of H2O in 480.0 g:
n(H2O) = 480.0 g / 18.015 g·mol−1 ≈ 26.662 mol

So n(H3PO4) = 26.662 / 3 = 8.887 mol

4) Mass of H3PO4 (M ≈ 97.99 g·mol−1):
m = 8.887 mol × 97.99 g·mol−1 ≈ 871 g

Answer: about 8.89 mol H3PO4, or ≈ 871 g H3PO4 (rounded).