Respuesta :
Answer: 45 grams
Explanation:
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
a) moles of [tex]Al[/tex]
[tex]\text{Number of moles}=\frac{82.49g}{27g/mol}=3.1moles[/tex]
b) moles of [tex]O_2[/tex]
[tex]\text{Number of moles}=\frac{117.65g}{32g/mol}=3.7moles[/tex]
[tex]4Al+3O_2\rightarrow 2Al_2O_3[/tex]
According to stoichiometry :
4 moles of [tex]Al[/tex] require = 3 moles of [tex]O_2[/tex]
Thus 3.1 moles of [tex]Al[/tex] will require=[tex]\frac{3}{4}\times 3.1=2.3moles[/tex] of [tex]O_2[/tex]
Thus [tex]Al[/tex] is the limiting reagent as it limits the formation of product and [tex]O_2[/tex] is the excess reagent as (3.7-2.3)= 1.4 moles of [tex]O_2[/tex] will remain unreacted.
Mass of [tex]O_2=moles\times {\text {Molar mass}}=1.4moles\times 32g/mol=45g[/tex]
Thus 45 g of [tex]O_2[/tex] will be present in the vessel when the reaction is complete.
