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19 September, 05:40

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.

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  1. 19 September, 06:56
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    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
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