Question

When $FeC{l}_3$ is ignited in an atmosphere of pure oxygen, the following reaction takes place:
$4FeC{l}_3\left(s\right)+3O_2\left(g\right)\to 2F{e}_2{O}_3\left(s\right)+6C{l}_2\left(g\right)$
If 3 moles of $FeC{l}_3$ are ignited in the presence of 2 moles of $O_2$ gas, how much of which reagent is present in excess and therefore, remains unreacted ?

• A. 0.33 mole $FeC{l}_3$ remains unreacted
• B. 0.67 mole $FeC{l}_3$ remains unreacted
• C. 0.25 mole $O_2$ remains unreacted
• D. 0.50 mole $O_2$ remains unreacted

Answer 1

The correct option is A 0.33 mole $FeC{l}_3$ remains unreacted

$4FeC{l}_3\left(s\right)+3O_2\left(g\right)\to 2F{e}_2{O}_3\left(s\right)+6C{l}_2\left(g\right)$.
The balanced equation shows that 4 moles of $FeC{l}_3$ reacts completely with 3 moles of $O_2$.
Hence 0.33 mole $FeC{l}_3$ remains unreacted.