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Sagot :
Answer: 65.28 g of [tex]Al_2O_3[/tex] will be produced from 34.6 g of Al.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Al=\frac{34.6g}{27g/mol}=1.28moles[/tex]
The balanced chemical reaction 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 1.28 moles of [tex]Al[/tex] produce=[tex]\frac{2}{4}\times 1.28=0.64moles[/tex] of [tex]Al_2O_3[/tex]
Mass of [tex][tex]Al_2O_3[/tex]= [tex]moles\times {\text {Molar mass}}=0.64moles\times 102g/mol=65.28g[/tex]
Thus 65.28 g of [tex]Al_2O_3[/tex] will be produced from 34.6 g of Al.
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