Question

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

NEVERMIND, I FIGURED IT OUT

Answer 1

Answer: 131 g of bromine is required.

Step-by-step 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}=(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=
`(3)/(2)* 0.544=0.816moles` of
`Br_2`

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

Thus 131 g of bromine is required.