# 2kbr(aq)+cl2(aq)>2kcl(aq)+br2(l) When 0.855g of cl2 and 3.205g of kbr are mixed in solution, which is the limiting reactant?

Apr 3, 2018

$C {l}_{2}$

#### Explanation:

The reaction
$2 K B r \left(a q\right) + C {l}_{2} \left(a q\right) \rightarrow 2 K C l \left(a q\right) + B {r}_{2} \left(l\right)$
shows that for every $C {l}_{2}$ molecule needed for the reaction, $2 K B r$ molecules are needed too.

For maximal yield and no excess reactants, the molar ratio $C {l}_{2} : K B r$ should be $1 : 2$.

However, when you calculate the moles of each reactant,
you get

$\left(.855 g C {l}_{2}\right) \cdot \frac{1 m o l C {l}_{2}}{70.9 g} = .01206 m o l C {l}_{2}$

$\left(3.205 g K B r\right) \cdot \frac{1 m o l K B r}{119.002 g} = .02693 m o l K B r$

The molar ratio $C {l}_{2} : K B r$ is $\frac{.02693}{.01206} = 2.233$.

That means there are more moles of $K B r$ needed to react with all of the $C {l}_{2}$ moles, so there is an excess of $K B r$.

Then it follows that $C {l}_{2}$ is the limiting reactant, since it is limiting the production of the reaction.