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Helga [31]
3 years ago
15

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

Chemistry
1 answer:
ankoles [38]3 years ago
5 0

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.

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