Question

If you had excess chlorine, how many moles of of aluminum chloride could be produced from 11.0 g of aluminum

Answer 1

Answer: 0.407 moles of aluminum chloride could be produced from 11.0 g of aluminum.

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of aluminium}=(11.0g)/(27g/mol)=0.407moles`

`2Al+3Cl_2\rightarrow 2AlCl_3`

According to stoichiometry :

`Al` is the limiting reagent as it limits the formation of product and
`Cl_2` is the excess reagent.

As 2 moles of
`Al` give = 2 moles of
`AlCl_3`

Thus moles of
`Al` give =
`(2)/(2)* 0.407=0.407moles` of
`AlCl_3`

Thus 0.407 moles of aluminum chloride could be produced from 11.0 g of aluminum.