Answer: 103.8 g of [tex]Fe_2O_3[/tex], 0.65 moles of
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} O_2=\frac{31.1}{32}=0.975moles[/tex]
Thus [tex]O_2[/tex] is the limiting reagent as it limits the formation of product and [tex]Fe[/tex] is the excess reagent.
[tex]4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3[/tex]
According to stoichiometry :
3 moles of [tex]O_2[/tex] produce = 2 moles of [tex]Fe_2O_3[/tex]
Thus 0.975 moles of [tex]O_2[/tex] will produce=[tex]\frac{2}{3}\times 0.975=0.65moles[/tex] of [tex]Fe_2O_3[/tex]
Mass of [tex]Fe_2O_3=moles\times {\text {Molar mass}}=0.65moles\times 159.69g/mol=103.8g[/tex]
Thus 103.8 g of [tex]Fe_2O_3[/tex] will be produced.
