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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:
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:
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.
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].