Balanced Equation
$"2Fe"_2"O"_3 + "3C"$$rarr$$"4Fe + 3CO"_2$
This is a limiting reactant question. The maximum amount of $"Fe"$ that can be produced is determined by the limiting reactant. We have to determine the amount of iron that can be produced by $"500"$ grams of iron(III) oxide, and $60$ grams of coke.
Mass of Iron Produced by Iron(III) oxide.
$color(red)("Convert the mass of iron(III) oxide to moles by multiplying by the inverse of its molar mass, 159.687 g/mol."$
$color(blue)("Determine the moles of iron produced by multiplying the moles iron by the mole ratio between iron and iron(III) oxide":$
$"4 mol Fe":$$"2 mol Fe"_2"O"_3"$
$color(green)"Determine the mass of iron produced by multiplying the moles iron by its molar mass"$.
$550"g Fe"_2"O"_3xxcolor(red)((1"mol Fe"_2)/(159.687"g Fe"_2"O"_3))xxcolor(blue)((4"mol Fe")/(2"mol Fe"_2"O"_3))xxcolor(green)((55.845"g Fe")/(1"mol Fe"))="385 g Fe"$
Mass of Iron Produced by Coke
Follow the same procedure as above, substituting $"C"$ for $"Fe"_2"O"_3"$, and use ratio $4"mol Fe:"$ $3"mol C"$.
$60"g C"xxcolor(red)((1"mol C")/(12.011"g C"))xxcolor(blue)((4"mol Fe")/(3"mol C"))xxcolor(green)((55.845"g Fe")/(1"mol Fe"))="372 g Fe"$
Coke is the limiting reactant because it produces less iron than iron(III) oxide.
I didn't round to the proper number of sig figs because it would give the wrong answer.
