Answer: 28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Al=\frac{15.0g}{27g/mol}=0.556moles[/tex]
The balanced chemical equuation is:
[tex]4Al+3O_2\rightarrow 2Al_2O_3[/tex]
According to stoichiometry :
4 moles of [tex]Al[/tex] produce == 2 moles of [tex]Al_2O_3[/tex]
Thus 0.556 moles of [tex]Al[/tex] will produce=[tex]\frac{2}{4}\times 0.556=0.278moles[/tex] of [tex]Al_2O_3[/tex]
Mass of [tex]Al_2O_3=moles\times {\text {Molar mass}}=0.278moles\times 102g/mol=28.4g[/tex]
Thus 28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal.