What volume of a 0.177 molL−1 Na3PO4 solution is necessary to completely react with 93.1 mL of 0.113 molL−1 CuCl2?

Question

the final answer has to be in ml

Answer 1

The first thing is to write a correctly balanced equation for the reaction:

2Na₃PO₄(aq) + 3CuCl₂(aq) ==> 6NaCl(aq) + Cu₃(PO₄)₂(s) ... balanced equation

Next, we find the moles of CuCl₂ that are present:

93.1 ml x 1 L/1000 ml x 0.113 mol/L = 0.01052 mols CuCl₂ present

Next, we use the stoichiometry of the balanced equation to find the moles of Na₃PO₄ needed to react with the moles of CuCl₂:

0.01052 mols CuCl₂ x 2 mols Na₃PO₄ / 3 mols CuCl₂ = 0.007014 molsNa₃PO₄ needed

Finally, we find the volume of Na₃PO₄ needed to provide this many moles of Na₃PO₄:

0.007014 mols x 1 L / 0.177 mols = 0.03962 L x 1000 ml / L = 39.6 mls Na₃PO₄ (3 sig. figs.)