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