# Question aca1d

Oct 4, 2015

$\text{699 g}$

#### Explanation:

The idea here is that you need to use the mole ratio that exists between ferric oxide, ${\text{Fe"_2"O}}_{3}$, and iron metal, $\text{Fe}$, to determine how many moles of the latter will be produced when all the given mass of the ferric oxide reacts.

${\text{Fe"_2"O"_text(3(s]) + 3"CO"_text((g]) -> color(purple)(2)"Fe"_text((s]) + 3"CO}}_{\textrm{2 \left(g\right]}}$

Notice that every mole of ferric oxide will produce $\textcolor{p u r p \le}{2}$ moles of iron metal.

To determine how many mole of ferric oxide you get in $\text{1.00 kg}$ of the compound, use its molar mass

1.00color(red)(cancel(color(black)("kg"))) * (1000color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * ("1 mole Fe"""_2"O"_3)/(159.69color(red)(cancel(color(black)("g")))) = "6.262 moles Fe"""_2"O"_3

Now use the 1:color(purple(2) mole ratio to find how many moles of iron metal will be produced by the reaction

6.262color(red)(cancel(color(black)("moles Fe"""_2"O"_3))) * (color(purple)(2)" moles Fe")/(1color(red)(cancel(color(black)("mole Fe"""_2"O"_3)))) = "12.524 moles Fe"

Finally, use iron's molar mass to find how many grams will contain this many moles of iron

12.524color(red)(cancel(color(black)("moles"))) * "55.845 g"/(1color(red)(cancel(color(black)("mole")))) = "699.4 g Fe"

Rounded to three sig figs, the answer will be

m_"Fe" = color(green)("699 g")#