Question

How many grams of Al2O3 can form from 34.6 grams of Al

Answer 1

Answer: 65.28 g of
`Al_2O_3` will be produced from 34.6 g of Al.

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} Al=(34.6g)/(27g/mol)=1.28moles`

The balanced chemical reaction is

`4Al+3O_2\rightarrow 2Al_2O_3`

According to stoichiometry :

4 moles of
`Al` produce = 2 moles of
`Al_2O_3`

Thus 1.28 moles of
`Al` produce=
`(2)/(4)* 1.28=0.64moles` of
`Al_2O_3`

Mass of
`[tex]Al_2O_3`=
`moles* {\text {Molar mass}}=0.64moles* 102g/mol=65.28g`

Thus 65.28 g of
`Al_2O_3` will be produced from 34.6 g of Al.