Answer : [tex]BaCl_2[/tex] reagent predict to be in excess.
Explanation : Given,
Concentration of [tex]BaCl_2[/tex] = 0.10 M
Volume of [tex]BaCl_2[/tex] = 7.50 mL = 0.0075 L (1 L = 1000 mL)
Concentration of [tex]KIO_3[/tex] = 0.10 M
Volume of [tex]KIO_3[/tex] = 7.50 mL = 0.0075 L
First we have to calculate the moles of [tex]BaCl_2[/tex] and [tex]KIO_3[/tex].
[tex]\text{Moles of }BaCl_2=\text{Concentration of }BaCl_2\times \text{Volume of }BaCl_2[/tex]
[tex]\text{Moles of }BaCl_2=0.10M\times 0.0075L=0.00075mol[/tex]
and,
[tex]\text{Moles of }KIO_3=\text{Concentration of }KIO_3\times \text{Volume of }KIO_3[/tex]
[tex]\text{Moles of }KIO_3=0.10M\times 0.0075L=0.00075mol[/tex]
Now we have to calculate the excess and limiting reagent.
The balanced equilibrium reaction will be:
[tex]BaCl_2+2KIO_3\rightleftharpoons Ba(IO_3)_2+2KCl [/tex]
From the balanced reaction we conclude that
As, 2 mole of [tex]KIO_3[/tex] react with 1 mole of [tex]BaCl_2[/tex]
So, 0.00075 moles of [tex]KIO_3[/tex] react with [tex]\frac{0.00075}{2}=0.000375[/tex] moles of [tex]BaCl_2[/tex]
From this we conclude that, [tex]BaCl_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]KIO_3[/tex] is a limiting reagent and it limits the formation of product.
Hence, [tex]BaCl_2[/tex] reagent predict to be in excess.
