Question

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?

Answer 1

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 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