Question

Question #57b23

Answer 1

`"1.3 g Fe"`

Explanation:

For starters, you need a balanced chemical equation to work with

`"Fe"_ 2"O"_ (3(s)) + color(blue)(3)"CO"_ ((g)) -> color(purple)(2)"Fe"_ ((s)) + 3"CO"_ (2(g))`

Now, you can use the molar masses of the two reactants and of elemental iron to say that for every

`1 color(red)(cancel(color(black)("mole Fe"_2"O"_3))) * "159.69 g"/(1color(red)(cancel(color(black)("mole Fe"_2"O"_3)))) = "159.69 g"`

of iron(III) oxide that take part in the reaction, the reaction consumes

`color(blue)(3) color(red)(cancel(color(black)("moles CO"))) * "28.01 g"/(1color(red)(cancel(color(black)("mole CO")))) = color(blue)("84.03 g")`

of carbon monoxide and produces

`color(purple)(2) color(red)(cancel(color(black)("moles Fe"))) * "55.845 g"/(1color(red)(cancel(color(black)("mole Fe")))) = color(purple)("111.69 g")`

of elemental iron. Now, use the mass of iron(III) oxide to see if you have enough carbon monoxide available to ensure that all the mass of iron(III) oxide takes part in the reaction.

So, you know that `"5 g"` of iron(III) oxide would require

`5 color(red)(cancel(color(black)("g Fe"_2"O"_3))) * color(blue)("84.03 g CO")/(159.69color(red)(cancel(color(black)("g Fe"_2"O"_3)))) = "2.631 g CO"`

Since you only have `"1 g"` of carbon monoxide available, you can say that carbon monoxide will act as a limiting reagent, i.e. it will be completely consumed before all the mass of iron(III) oxide will get the chance to react.

Consequently, the reaction will consume `"1 g"` of carbon monoxide and

`1 color(red)(cancel(color(black)("g CO"))) * ("159.69 g Fe"_2"O"_3)/(color(blue)(84.03)color(red)(cancel(color(blue)("g CO")))) = "1.900 g Fe"_2"O"_3`

and produce

`1 color(red)(cancel(color(black)("g CO"))) * color(purple)("111.69 g Fe")/(color(blue)(84.03)color(red)(cancel(color(blue)("g CO")))) = color(darkgreen)(ul(color(black)("1.3 g Fe")))`

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for your values.