Question

Question #aca1d

Answer 1

`"699 g"`

Explanation:

The idea here is that you need to use the mole ratio that exists between ferric oxide, `"Fe"_2"O"_3`, and iron metal, `"Fe"`, to determine how many moles of the latter will be produced when all the given mass of the ferric oxide reacts.

`"Fe"_2"O"_text(3(s]) + 3"CO"_text((g]) -> color(purple)(2)"Fe"_text((s]) + 3"CO"_text(2(g])`

Notice that every mole of ferric oxide will produce `color(purple)(2)` moles of iron metal.

To determine how many mole of ferric oxide you get in `"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")`