Answer: 73.4 g of [tex]Fe_2O_3[/tex] are formed when 51.3 grams of iron react completely with oxygen.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Fe=\frac{51.3g}{56g/mol}=0.92moles[/tex]
The balanced chemical reaction is:
[tex]4Fe+3O_2\rightarrow 2Fe_2O_3[/tex]
As [tex]Fe[/tex] is the limiting reagent as it limits the formation of product.
According to stoichiometry :
4 moles of [tex]Fe[/tex] produce = 2 moles of [tex]Fe_2O_3[/tex]
Thus 0.92 moles of [tex]Fe[/tex] will produce=[tex]\frac{2}{4}\times 0.92=0.46moles[/tex] of [tex]Fe_2O_3[/tex]
Mass of [tex]Fe_2O_3=moles\times {\text {Molar mass}}=0.46moles\times 159.69g/mol=73.4g[/tex]
Thus 73.4 g of [tex]Fe_2O_3[/tex] are formed when 51.3 grams of iron react completely with oxygen.
