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

Step-by-step 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 * 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 * (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 =
`(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.