Question

What volume of a 0.2089 M KI solution contains enough KI to react exactly with the Cu(NO₃)₂ in 43.88 mL of a 0.3842 M solution of Cu(NO₃)₂?

2Cu(NO₃)₂+4KI⟶2CuI+I₂+4KNO₃.
a) 21.94 mL
b) 43.88 mL
c) 65.82 mL
d) 87.76 mL

Answer 1

To find the volume of KI solution needed to react with Cu(NO₃)₂, the moles of Cu(NO₃)₂ are calculated and then used to determine the moles of KI needed based on the stoichiometric ratio. After calculating, the required volume of 0.2089 M KI solution is 161.4 mL, which rounds to 160 mL. However, none of the given answer choices match this calculation.

Step-by-step explanation:

To determine the volume of a 0.2089 M KI solution needed to react with 43.88 mL of a 0.3842 M Cu(NO₃)₂ solution, we look at the stoichiometry of the balanced chemical equation:

2Cu(NO₃)₂ + 4KI → 2CuI + I₂ + 4KNO₃.

This equation shows that 2 moles of Cu(NO₃)₂ react with 4 moles of KI. Therefore, there is a 1:2 molar ratio between Cu(NO₃)₂ and KI.

First calculate the moles of Cu(NO₃)₂:

1. (0.04388 L) × (0.3842 mol/L) = 0.01686 mol of Cu(NO₃)₂.

Then, use the molar ratio to find moles of KI required:

1. (0.01686 mol Cu(NO₃)₂) × (2 mol KI / 1 mol Cu(NO₃)₂) = 0.03372 mol KI.

Finally, calculate the volume of KI solution:

1. Volume KI = moles KI / Molarity of KI = 0.03372 mol / 0.2089 M = 0.1614 L or 161.4 mL.

To select the best answer from the provided options, we round to two significant figures because the molarity of KI was given to four significant figures:

161.4 mL rounds to 160 mL, which is not an option. The closest listed option is 87.76 mL (d), which is not accurate. There may be a mistake in the question or the provided options. Hence, none of the options (a) 21.94 mL, (b) 43.88 mL, (c) 65.82 mL, or (d) 87.76 mL accurately reflect the required volume of KI solution.