Question

The number of moles of $F{e}_2{O}_3$ formed when $5.6$ of $O_2$ reacts with $5.6gm$ of $Fe$ is :

• A. $0.125$
• B. $0.01$
• C. $0.05$
• D. $0.1$

Answer 1

The correct option is C $0.05$

Balanced Reaction is:

$4Fe+3O_2\to 2F{e}_2{O}_3$

$5.6gm$ $5.6gm$

$\left(\frac{5.6}{56}\right)$ $\left(\frac{5.6}{32}\right)$

$0.1mole0.175mole$

$4molesFe$ reacts with $3moles$ of $O_2$ .

$\therefore$ 0.1 mole $Fe$ reacts with $\frac{3\times 0.1}{4}moles{O}_2$

Here, Fe is limiting reagent.

4 moles of $Fe$ form 2 moles of $F{e}_2{O}_3$

Mole of $F{e}_2{O}_3$ formed by 0.1 mole$Fe=\frac{2}{4}\times 0.10$

$=0.05mole$

The correct option is ${}^{\prime }{C}^{\prime }$.