Question

Amount of the excess reagent left at the end of the reaction is :
(Given that molecular weight of $Al=27,Fe=56,O=16$ g/mol)

• A. $1.46$ mol
• B. $1.36$ mol
• C. $2.35$ mol
• D. none of these

Answer 1

The correct option is A $1.46$ mol

The reaction is as follows:
$2Al+F{e}_2{O}_3⟶A{l}_2{O}_3+2Fe$
$2$ $1$
$124$ g $601$ g
$\frac{124}{27}=4.6$ mol $\frac{601}{160}=3.76$ mol
$4.6$ $3.76-2.3=1.46$ mol left
Hence, excess reagent $\left(F{e}_2{O}_3\right)$ left = $1.46$ mol.