Question #aca1d

1 Answer
Oct 4, 2015

Answer:

#"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")#