Question

How many grams of bromine are required to react completely with 37.4 grams aluminum chloride?

AlCl3 + Br2 → AlBr3 + Cl2

Answer 1

The balanced chemical equation is

`2AlCl_3  +3 Br_2  \Rightarrow 2 AlBr_3  + 3Cl_2`

The conversions are

mass
`AlCl_3` to moles
`AlCl_3` to moles
`Br_2` to mass
`Br_2`

To convert mass
`AlCl_3` to moles
`AlCl_3` we divide the mass by molar mass
`AlCl_3` which is

`26.9+(3*35.5)=133.34 g/mol`

To convert moles
`AlCl_3` to moles
`Br_2` we make use of mole ratio 2 : 3

To convert moles
`Br_2` to mass
`Br_2` we multiply by molar mass
`Br_2` which is

`80 * 2=160 (g )/mol`

`$37.4 g A l C l_(3) * (1 m o l A l C l_(3))/(133.34 g A l C l_(3)) * (3 m o l B r_(2))/(2 m o l A l C l_(3)) * (160 g B r_(2))/(1 m o l B r_(2))$`

=67.2g
`Br_2` is required (Answer).