When NaI is added to a solution of Cu(NO₃)₂, the following reaction occurs: 2NaI (aq) → CuI₂ (s) + 2NaNO₃ (aq)

If a chemist needs to make 350.0 g of CuI₂, how many mL of a 3.3 M solution of NaI must be used? Assume that there is excess copper (II) nitrate.

668 mL is the volume of NaI that must be added.

Explanation:

The reaction is:

Cu(NO₃)₂ (aq)  +  2NaI (aq)  →  CuI₂ (s) + 2NaNO₃ (aq)

If the copper(II) nitrate is in excess, the limiting reactant is the iodide

2 moles of sodium iodide produce 1 mol of copper(II) iodide.

I made 350 g of CuI₂, so I made (350 g . 1 mol/317.35 g) = 1.10 moles

Rule of three will be:

1 mol of CuI₂ was produced by 2 moles of NaI

1.10 mol of CuI₂ must been produced ny (1.10 . 2) /1 = 2.20 moles of NaI

Molarity = mol/L = mol / volume → Therefore,  volume = mol/ molarity

2.20 mol / 3.30 mol/L = 0.668 L → 668 mL