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