# 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 9, 2017

188 g of Potassium Chloride

#### Explanation:

This is a limiting reactant problem so we need to see which will produce the least amount of moles of potassium chloride then use that to find how many grams.

$300.0 g C {l}_{2} \frac{1 m o l C {l}_{2}}{70.906 g} = 4.231 m o l C {l}_{2} \frac{2 m o l K C l}{1 m o l C {l}_{2}} = 8.462 m o l K C l$

$300.0 g K B r \frac{1 m o l K B r}{119.002 g} = 2.521 m o l K B r \frac{2 m o l K C l}{2 m o l K B r} = 2.521 m o l K C l$

The limiting reactant is KBr from this we will determine the amount of potassium chloride.

$2.521 m o l K C l \left(74.5513 \frac{g}{m o l}\right) = 187.941 g$

To the correct number of sig figs this is rounded to 188 g