Question

Aluminum and cloride undergoes a synthesis reaction if 8molnof Al reacts with 10mol of cl, what is the maximum amount of AlCl3 can produce

Answer 1

Answer: The mass of
`AlCl_3` produced is 889.38 g

Step-by-step explanation:

We are given:

Moles of Al = 8 mol

Moles of
`Cl_2` = 10 mol

For the given chemical reaction:

`2Al+3Cl_2\rightarrow 2AlCl_3`

By stoichiometry of the reaction:

If 3 moles of chlorine gas reacts with 2 moles of Al

So, 10 moles of chlorine gas will react with =
`(2)/(3)* 10=6.67mol` of Al

As the given amount of Al is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, chlorine gas is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 3 moles of
`Cl_2` produces 2 mole of
`AlCl_3`

So, 10 moles of
`Cl_2` will produce =
`(2)/(3)* 10=6.67mol` of
`AlCl_3`

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

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

We know, molar mass of
`AlCl_3` = 133.34 g/mol

Putting values in above equation, we get:

`\text{Mass of }AlCl_3=(6.67mol* 133.34g/mol)=889.38g`

Hence, the mass of
`AlCl_3` produced is 889.38 g