Question

What volume of a 0.2089 M KI solution contains enough KI to react exactly with the Cu(NO3)2 in 40.88 mL of a 0.3842 M solution of Cu(NO3)2
2Cu(NO3)2 +4KI ? 2CuI + I2 + 4KNO3

Answer 1

Explanation:

As molarity is the number of moles placed in a liter of solution. Therefore, no. of mole = Molarity × volume of solution in liter

Hence, moles of $Cu(NO_{3})_{2}$ will be calculated as follows.

No. of mole of $Cu(NO_{3})_{2}$ = $0.3842 M \times 0.04388 L$ = 0.0168 mole

According to the given reaction, 2 mole of $Cu(NO_{3})_{2}$ react with 4 mole of KI.

Therefore, for 0.0168 mole amount of $Cu(NO_{3})_{2}$ required will be as follows.

$Cu(NO_{3})_{2}$ = $0.0168 \times \frac{4}{2}$

                       = 0.0337 mole of KI

Hence, volume of KI required will be calculated as follows.

               Volume = $\frac{\text{no. of moles}}{Molarity}$

    Volume of KI = $\frac{0.0337}{0.2089}$

                           = 0.1614 liter

                            = 161.4 ml            (as 1 L = 1000 mL)

Thus, we can conclude that 161.4 ml  volume of a 0.2089 M KI is required for the given situation.