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?

1 Answer
Apr 9, 2017


188 g of Potassium Chloride


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 Cl_2(1 mol Cl_2) /(70.906 g) = 4.231 mol Cl_2(2 mol KCl)/(1 mol Cl_2)=8.462 mol KCl#

#300.0 g KBr(1 mol KBr ) /(119.002 g) = 2.521 mol KBr(2 mol KCl)/(2 mol KBr)=2.521 mol KCl#

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

#2.521 mol KCl(74.5513 g/(mol)) = 187.941 g #

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