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.

Explanation:

To calculate the moles :

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

$\text{Moles of} Al=\frac{34.6g}{27g/mol}=1.28moles$

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=$\frac{2}{4}\times 1.28=0.64moles$  of $Al_2O_3$

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

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