# For the reaction represented by the equation Cl_2 + 2KBr -> 2KCl + Br_2, how many grams of potassium chloride can be produced from 300. g each of chlorine and potassium bromide?

Apr 13, 2017

#### Answer:

See the explanationion below$:$

#### Explanation:

$C {l}_{2}$ + $2 K B r$ $\to$ $2 K C l$ + $B {r}_{2}$
Given:
$300 g$ of $C {l}_{2}$
$300 g$ of $K B r$

(a) Mass of $C {l}_{2}$ = 300g
Molecular mass of $C {l}_{2}$ = 2(35.45) = 70.906 g$/$$m o l$
No.of moles of $C {l}_{2}$ = $m a s s$$/$$m o l a r m a s s$ = $300$$/$$70.906$
=$4.23$ moles
Mass of $K B r$ = $300 g$
Molar mass of $K B r$= $39 + 79.9$ = $118.9$$g$$/$$m o l$
No.of moles of $K B r$ = $300$$/$$118.9$
=$2.52$moles
Now,
Consider the equation:
$C {l}_{2}$ +$2 K B r$$\to$ $2 K C l$ + $B {r}_{2}$
$1$ mole of $C {l}_{2}$ = 2 moles of $K C l$
$4.23$ moles of $C {l}_{2}$ = 2x4.23= $8.46$ moles

$2$ moles of $K B r$ = 2 moles of $K C l$
$2.52$ moles of $K B r$ = $2.52$ moles

Now convert moles to mass,
Mass of $K C l$ produced by $C {l}_{2}$ = 8.46×70.906=$599.8 g$
Mass of $K C l$ produced by potassium bromide = 2.52×74.5=187.74g

As $K B r$ is producing lesser moles of product so, $K B r$ is limiting reactant and produced potassium chloride= $187.74 g$

Apr 15, 2017

#### Answer:

They can be produced about 188 grams of KCl.

#### Explanation:

At first, I calculated the mole amount of Cl₂

$\text{300 g"/"35.45 g/mol" = 8.463\ mol}$

and KBr:

$\text{300 g"/"119.01 g/mol} = 2.521 \setminus m o l$.

Then, by observing that half of dichlorine moles (4.232 mol) are greater than the available moles of KBr, I deduced potassium bromide is the limiting reactant.

Consequently, provided the stoichiometric coefficient of the required product, KCl, is the same of KBr, 2.521 moles of KBr will be converted in the same mole amount of KCl.

Eventually, I calculated the mass of 2.521 moles of KCl:

2.521 mol · 74.55 g/mol = 187.9 g.

(by student Alessia, checked by prof. a.t.)