Answer: a) [tex]Fe_2O_3[/tex]
b) 92.1 g
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]Fe_2O_3[/tex]
[tex]\text{Number of moles}=\frac{1.873\times 10^2 g}{159.69g/mol}=1.173moles[/tex]
b) moles of [tex]Al[/tex]
[tex]\text{Number of moles}=\frac{94.51g}{27g/mol}=3.500moles[/tex]
[tex]Fe_2O_3+2Al\rightarrow Al_2O_3+2Fe[/tex]
According to stoichiometry :
1 mole of [tex]Fe_2O_3[/tex] require 2 moles of [tex]Al[/tex]
Thus 1.173 [tex]Cl_2[/tex] require=[tex]\frac{2}{1}\times 1.173=2.346moles[/tex] of [tex]Al[/tex]
Thus [tex]Fe_2O_3[/tex] is the limiting reagent as it limits the formation of product. Thus [tex]Fe_2O_3[/tex] will run out first.
[tex]Al[/tex] is the excess reagent as (3.500-2.346)= 1.154 moles are left unreacted.
2 moles of [tex]Al[/tex] require 1 mole of [tex]Fe_2O_3[/tex]
3.500 moles of [tex]Al[/tex] require= [tex]\frac{1}{2}\times 3.500=1.750moles[/tex] of [tex]Fe_2O_3[/tex]
Moles of [tex]Fe_2O_3[/tex] required = (1.750-1.173) = 0.577
Mass of [tex]Fe_2O_3=moles\times {\text {Molar mass}}=0.577moles\times 159.69g/mol=92.1g[/tex]
Thus he should order 92.1 g of the limiting reactant.
