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.

Step-by-step 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=
`(1)/(1)* 2=2moles` of
`MgCl_2`

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

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