Answer:
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
