Question

Based on the equation, how many grams of Br2 are required to react completely with 72.4 grams of AlCl3

Answer 1

Answer:  131 g of bromine is required.

Explanation:

The balanced equation will be :

$2AlCl_3+3Br_2\rightarrow 2AlBr_3+3Cl_2$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

moles of $AlCl_3$

$\text{Number of moles}=\frac{72.4g}{133g/mol}=0.544moles$

According to stoichiometry :

2 moles of $AlCl_3$ require  = 3 moles of $Br_2$

Thus 0.544 moles of $AlCl_3$ require=$\frac{3}{2}\times 0.544=0.816moles$  of $Br_2$

Mass of $Br_2=moles\times {\text {Molar mass}}=0.816moles\times 160g/mol=131g$

Thus 131 g of bromine is required.