##”699 g”##

The idea here is that you need to use the 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 , the answer will be

##m_”Fe” = color(green)(“699 g”)##