Question

Which of the following statement(s) is/are incorrect when 0.15 mol of $Fe$ is allowed to react with $4.8\text{g}$ of $O_2$?
(Molar mass of $Fe=56\text{g/mol}$)
$4Fe\left(s\right)+3O_2\left(g\right)\to 2F{e}_2{O}_3\left(s\right)$

• A. $Fe$ is the limiting reagent
• B. The mass of $O_2$ left over at the end of the reaction is $1.2\text{g}$
• C. The mass of $F{e}_2{O}_3$ produced is $12.0\text{g}$
• D. $O_2$ is the limiting reagent

Answer 1

The correct option is D $O_2$ is the limiting reagent

$4Fe+3O_2\to 2F{e}_2{O}_3$
Moles of $O_2=\frac{4.8}{32}=0.15$ mol

Finding the limiting reagent,
$\text{For}Fe=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{0.15}{4}=0.0375$
$\text{For}{O}_2=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{0.15}{3}=0.05$

So, $Fe$ is the limiting reagent.

4 mol of $Fe$ react with 3 mol of $O_2$
0.15 mol of $Fe$ will react with, $\frac{3}{4}\times 0.15=0.1125$ moles of $O_2$

Amount of $O_2$ reacted = $0.1125\times 32=3.6\text{g}$
Amount of $O_2$ left = $4.8-3.6=1.2\text{g}$

4 moles of $Fe$ produces 2 moles of $F{e}_2{O}_3$
Molar mass of $F{e}_2{O}_3=56\times 2+3\times 16=160\text{g/mol}$
0.15 mol will produce $\frac{2\times 0.15}{4}$ = 0.075 mol of $F{e}_2{O}_3$

Amount of $F{e}_2{O}_3=0.075\times 160=12\text{g}$
$O_2$ is the excess reagent.
Therefore, option (d) is the incorrect statement.