3 Cu(OH)2 + 2 H3PO4 --> Cu3(PO4)2 + 6 H2O

Question

I know that the given is 2.14 moles of copper (II) hydroxide and what its asking to find is grams of Cu3(PO4)2 anytime I try to get the answer it doesn't come out correctly.

If 2.14 moles of copper (II) hydroxide reacts with excess phosphoric acid to form water and copper(II) phosphate, how many grams of copper (II) phosphate will form?

Answer 1

In this stoichiometry, first convert moles of the reactant to moles of product using the coefficients in the balanced reaction, then multiply by molar mass of the product. It's basically dimensional analysis.

[2.14 mole Cu(OH)₂] * [1 mole Cu₃(PO₄)₂ / 3 mole Cu(OH)₂] * [380.58 g Cu₃(PO₄)₂ / 1 mole Cu₃(PO₄)₂] = 271 mole Cu₃(PO₄)₂.