Answered

Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Connect with a community of experts ready to provide precise solutions to your questions quickly and accurately. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

The reaction of iron (III) oxide with carbon monoxide produces iron and carbon dioxide.
Fe,O3(s) + 3CO(g) - 2Fe(s) + 3C0 (9)
If you have 39.5g of Fe2O3, how many grams of CO are required for a complete reaction?


Sagot :

mvschx

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

jacob193

Answer:

Approximately [tex]20.8\; \rm g[/tex].

Explanation:

[tex]\rm Fe_2O_3 \, (s) + 3\; CO\, (g) \to 2\; Fe\, (s) + 3\; CO_2\, (g)[/tex].

Relative atomic mass:

  • [tex]\rm Fe[/tex]: [tex]55.845[/tex].
  • [tex]\rm C[/tex]: [tex]12.011[/tex].
  • [tex]\rm O[/tex]: [tex]15.999[/tex].

Formula mass:

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

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

Number of moles of [tex]\rm Fe_2O_3[/tex] formula units in [tex]39.5\; \rm g[/tex] of this compound:

[tex]\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}[/tex].

Refer to the balanced equation for this reaction.

  • Coefficient of [tex]\rm Fe_2O_3[/tex]: [tex]1[/tex].
  • Coefficient of [tex]\rm CO[/tex]: [tex]3[/tex].

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

[tex]\displaystyle \frac{n({\rm CO})}{n({\rm Fe_2O_3})} = \frac{3}{1} = 3[/tex].

Calculate the number of moles of [tex]\rm CO[/tex] required to react with that [tex]39.5\; \rm g[/tex] of [tex]\rm Fe_2O_3[/tex]:

[tex]\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}[/tex].

Make use of the formula mass of [tex]\!\rm CO[/tex] to find the mass of that [tex]0.742\; \rm mol[/tex] of [tex]\rm CO[/tex] molecules:

[tex]\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}[/tex].

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.