Question

4. What reagent would you predict to be in excess for reacting 7.50 mL of a 0.10M BaCl2 solution with 7.50 mL of 0.10M KIO3 solution

Answer 1

Answer :
`BaCl_2` reagent predict to be in excess.

Explanation : Given,

Concentration of
`BaCl_2` = 0.10 M

Volume of
`BaCl_2` = 7.50 mL = 0.0075 L (1 L = 1000 mL)

Concentration of
`KIO_3` = 0.10 M

Volume of
`KIO_3` = 7.50 mL = 0.0075 L

First we have to calculate the moles of
`BaCl_2` and
`KIO_3`.

`\text{Moles of }BaCl_2=\text{Concentration of }BaCl_2* \text{Volume of }BaCl_2`

`\text{Moles of }BaCl_2=0.10M* 0.0075L=0.00075mol`

and,

`\text{Moles of }KIO_3=\text{Concentration of }KIO_3* \text{Volume of }KIO_3`

`\text{Moles of }KIO_3=0.10M* 0.0075L=0.00075mol`

Now we have to calculate the excess and limiting reagent.

The balanced equilibrium reaction will be:

`BaCl_2+2KIO_3\rightleftharpoons Ba(IO_3)_2+2KCl`

From the balanced reaction we conclude that

As, 2 mole of
`KIO_3` react with 1 mole of
`BaCl_2`

So, 0.00075 moles of
`KIO_3` react with
`(0.00075)/(2)=0.000375` moles of
`BaCl_2`

From this we conclude that,
`BaCl_2` is an excess reagent because the given moles are greater than the required moles and
`KIO_3` is a limiting reagent and it limits the formation of product.

Hence,
`BaCl_2` reagent predict to be in excess.