Question

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

Answer 1

Answer:

None  

Explanation:

Cl₂ is above Br₂ in the activity series.

Bromine will not displace chlorine from its salts.

The reaction will not occur.

Answer 2

Answer:

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

$26.9+(3\times35.5)=133.34 g/mol$

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

$80 \times 2=160 (g )/mol$

$37.4 g A l C l_{3} \times \frac{1 m o l A l C l_{3}}{133.34 g A l C l_{3}} \times \frac{3 m o l B r_{2}}{2 m o l A l C l_{3}} \times \frac{160 g B r_{2}}{1 m o l B r_{2}}$

=67.2g $Br_2$  is required (Answer).