Question

. How many grams of magnesium chloride can be produced by reacting 2 moles of chlorine gas with excess magnesium bromide? ____Cl2 + ____MgBr2  ___MgCl2 + ___Br2

Answer 1

Answer: 190 g of magnesium chloride can be produced by reacting 2 moles of chlorine gas with excess magnesium bromide.

Explanation:

The balanced chemical reaction is;

$Cl_2+MgBr_2\rightarrow MgCl_2+Br_2$

$Cl_2$ is the limiting reagent as it limits the formation of product and $MgBr_2$ is the excess reagent.

According to stoichiometry :

1 mole of $Cl_2$ produces = 1 mole of $MgCl_2$

Thus 2 moles of $Cl_2$ will produce=$\frac{1}{1}\times 2=2moles$  of $MgCl_2$

 Mass of $MgCl_2=moles\times {\text {Molar mass}}=2moles\times 95g/mol=190g$

Thus 190 g of magnesium chloride can be produced by reacting 2 moles of chlorine gas with excess magnesium bromide