Question

Which of the following statements is/are correct for the reaction given below?
$4Fe+3O_2\to 2F{e}_2{O}_3$

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

Answer 1

Answer: (A), (B), (C)

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

Moles of $O_2=\frac{4.8}{32}=0.15mol{O}_2$.
Moles of $Fe$ required $=\frac{4molFe}{3mol{O}_2}\times 0.15=0.2mol$.
a. Given mol of $Fe$ $=0.15$.
Hence, $Fe$ is the limiting reagent and no $Fe$ will remain after the reaction (Limiting reagent).
b. Weight of $O_2$ required $=\left(0.15molFe\right)$.
$\left(\frac{3mol{O}_2}{4molFe}\right)\left(\frac{32g{O}_2}{mol{O}_2}\right)$ $=\frac{0.15\times 3\times 32}{4}$ $=3.6g{O}_2$ required.
Weight of $O_2$ in excess $=\left(48O_{2}present\right)-\left(3.6g{O}_{2}required\right)$ $=1.2g{O}_2$ in excess
c. Weight of $F{e}_2{O}_3$ produced $=\left(0.15molFe\right)$.
$\left(\frac{2molF{e}_2{O}_3}{4molFe}\right)\left(\frac{160gF{e}_2{O}_3}{molF{e}_2{O}_3}\right)$ $=\frac{0.15\times 2\times 160}{4}$ $=12.0gF{e}_2{O}_3$ produced.
d. $O_2$ is not the limiting reagent.
Hence, options A,B & C are correct.