Question

What reagent would you predict to be in excess for reacting 6.25 mL of a 0.10 M of barium chloride solution with 6.25 mL of 0.10 M KIO3 solution?

Answer 1

Task:

What reagent would you predict to be in excess for reacting 6.25 mL of a 0.10 M of barium chloride solution with 6.25 mL of 0.10 M KIO3 solution?

Solution:

The chemical equation for this reaction is

$$\mathrm{BaCl_2} + 2\mathrm{KIO_3} = \mathrm{Ba(IO_3)_2} + 2\mathrm{KCl}$$

According to the equation the number of moles of $\mathrm{KIO_3}$ is twice the number of moles of $\mathrm{BaCl_2}$.

$$n(\mathrm{KIO_3}) = 2 \cdot n(\mathrm{BaCl_2})$$

The number of moles of the reactants is

$$n\ (\mathrm{mol}) = C(\mathrm{M}) \cdot V(\mathrm{L})$$ $$n\ (\mathrm{BaCl_2}) = 0.10 \cdot 6.25 \cdot 10^{-3} = 6.25 \cdot 10^{-4}\ \mathrm{mol}$$ $$n\ (\mathrm{KIO_3}) = 0.10 \cdot 6.25 \cdot 10^{-3} = 6.25 \cdot 10^{-4}\ \mathrm{mol}$$

According to the equation the number of moles of $\mathrm{BaCl_2}$ has to be

$$n\ (\mathrm{BaCl_2}) = 6.25 \cdot 10^{-4} / 2 = 3.125 \cdot 10^{-4}\ \mathrm{mol}$$

That is less than we have in solution, that's why $\mathrm{BaCl_2}$ is in excess.

Answer: $\mathrm{BaCl_2}$ is in excess