Question

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?

Answer 1

`Cl_2`

Explanation:

The reaction
`2KBr (aq) + Cl_2 (aq) rarr 2KCl (aq) + Br_2 (l)`
shows that for every `Cl_2` molecule needed for the reaction, `2KBr` molecules are needed too.

For maximal yield and no excess reactants, the molar ratio `Cl_2:KBr` should be `1:2`.

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

`(.855 g Cl_2)*(1 mol Cl_2)/(70.9g) = .01206 mol Cl_2`

`(3.205 g KBr)*(1 mol KBr)/(119.002 g) = .02693 mol KBr`

The molar ratio `Cl_2:KBr` is `.02693/.01206 = 2.233`.

That means there are more moles of `KBr` needed to react with all of the `Cl_2` moles, so there is an excess of `KBr`.

Then it follows that `Cl_2` is the limiting reactant, since it is limiting the production of the reaction.