Answer
$mass(Fe) = 244.62g$
$mass(CO_2) = 289.08g$
Work Step by Step
Balanced equation: (3.64a)
$Fe_2O_3+3CO−−>2Fe+3CO_2$
1. Convert the mass of (Fe2O3) to nº of moles:
$mm(Fe_2O_3)=55.85∗2+16∗3=159.7g/mol$
$n(moles)=mass(g)*mm$
$n(moles)=350*159.7$
$n(moles)=2.19$
2. Use the proportions of the balanced equation to find the number of moles of Fe and CO2 formed:
Fe:
$\frac{1}{2}=\frac{2.19}{x}$
$x=2∗2.19=4.38moles$
CO2:
$\frac{1}{3}=\frac{2.19}{y}$
$y=3∗2.19=6.57moles$
3. Convert these numbers to grams:
Fe:
$mm(Fe)=55.85g/mol$
$mass(g)=mm∗n(moles)$
$mass(g)=55.85∗4.38$
$mass(g)=244.62g$
CO2:
$mm(CO2)=12∗1+16∗2=44g/mol$
$mass(g)=mm∗n(moles)$
$mass(g)=44∗6.57$
$mass(g)=289.08g$
