Question

The reaction of iron (III) oxide with carbon monoxide produces iron and carbon dioxide.

Answer 1

Answer:

21g

Explanation:

no.ofmol fe2o3=39.5/(56×2+16×3)=0.25mol

from equation 1mole fe2o3 react with 3mole co

so,0.25mol fe2o3 react with 0.75mol co

mass of co=0.75×(12+16)=21g

Answer 2

Answer:

Approximately $20.8\; \rm g$.

Explanation:

$\rm Fe_2O_3 \, (s) + 3\; CO\, (g) \to 2\; Fe\, (s) + 3\; CO_2\, (g)$.

Relative atomic mass:

• $\rm Fe$: $55.845$.
• $\rm C$: $12.011$.
• $\rm O$: $15.999$.

Formula mass:

$\begin{aligned}M({\rm Fe_2O_3}) &= 2 \times 55.845 + 3 \times 15.999\\ &= 159.687\; \rm g \cdot mol^{-1}\end{aligned}$.

$\begin{aligned}M({\rm CO}) &= 12.011 + 15.999\\ &= 28.010\; \rm g \cdot mol^{-1}\end{aligned}$.

Number of moles of $\rm Fe_2O_3$ formula units in $39.5\; \rm g$ of this compound:

$\begin{aligned}&n({\rm Fe_2O_3}) \\ &= \frac{m({\rm Fe_2O_3})}{M({\rm Fe_2O_3})} \\ &= \frac{39.5\; \rm g}{159.687\; \rm g \cdot mol^{-1}}\approx 0.247\; \rm mol\end{aligned}$.

Refer to the balanced equation for this reaction.

• Coefficient of $\rm Fe_2O_3$: $1$.
• Coefficient of $\rm CO$: $3$.

Hence, for every formula unit of $\rm Fe_2O_3$ that this reaction consumes, $3\; \rm mol$ of $\rm CO$ molecules would also need to be consumed. Therefore, if neither reactant is in excess:

$\displaystyle \frac{n({\rm CO})}{n({\rm Fe_2O_3})} = \frac{3}{1} = 3$.

Calculate the number of moles of $\rm CO$ required to react with that $39.5\; \rm g$ of $\rm Fe_2O_3$:

$\begin{aligned}&n({\rm CO}) \\ &= n({\rm Fe_2O_3}) \cdot \frac{n({\rm CO})}{n({\rm Fe_2O_3})} \\[0.5em] &\approx 0.247\; \rm mol \times 3 \approx 0.742\; \rm mol\end{aligned}$.

Make use of the formula mass of $\!\rm CO$ to find the mass of that $0.742\; \rm mol$ of $\rm CO$ molecules:

$\begin{aligned} m({\rm CO}) &= n({\rm CO}) \cdot M({\rm CO}) \\ &\approx 0.742\; \rm mol \times 28.010\; \rm g \cdot mol^{-1} \\ &\approx 20.8\; \rm g\end{aligned}$.