For the reaction;

$C{l}_2\left(g\right)+KBr\left(s\right)\to KCl\left(s\right)+B{r}_2\left(g\right)$

How many grams of potassium chloride can be produced from 300 grams of potassium bromide? 

The given reaction is not balanced. The number of atoms in both sides are not equal.Consider balanced reaction;

$C{l}_2\left(g\right)+\text{ 2}KBr\left(s\right)\to 2KCl\left(s\right)+B{r}_2\left(g\right)$
From two mole of potassium bromide, two mole potassium chloride is produced. In terms of molar mass, from 119 g KBr, 74.5 g KCl is formed. Given; 300 g potassium bromide as reactant. Then the number of moles of KBr is;

$\frac{Given mass}{molar mass}=\frac{300}{119}=2.52 mols$

2.52 moles of KCl is also formed. The mass of KCl formed is;

$no. of mols\times molar mass=2.52\times 74.5=187.8\approx 188 g$

188 g of potassium chloride is formed.