(Checking Solution to problem) Using limiting reactants and known mass to determine mass of product?

"The equation for one of the reactions in the process of turning iron ore into the
metal is
Fe2O3(s) + 3 CO(g) -> 2 Fe(s) + 3 CO2(g)
If you start with 2.00 kg of each reactant, what is the maximum mass of iron you
can produce? "

At 2000g, 17 moles of Fe2O3 will require 1428g CO to fully react, meaning Fe2O3 should be my limiting reagent. Since I can see the mole ratio in the equation is 1:2 (Fe2O3:Fe) is it as simple as doubling the number of moles of Fe2O3 and multiplying by the molar weight of Fe? My answer following this method is 1904g Fe.

1 Answer
Sep 10, 2016


Here's my take on this.


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

I'm not really sure I follow your calculations on this one.

For starters, a very quick way of figuring out which reactant will act as the limiting reagent when dealing with equal masses of the two reactants is to look at their molar masses.

When you're dealing with equal masses of two reactants, the reactant that has the greater molar mass will have the smallest number of moles present in the given mass.

In this case, iron (III) oxide has a molar mass of #"159.7 g mol"^(-1)# and carbon monoxide has a molar mass of #"28.01 g mol"^(-1)#.

Right from the start, you should be able to tell that iron(III) oxide will be the limiting reagent because you have significantly more moles of carbon monoxide in #"2.00 kg"# than moles of iron(III) oxide in #"2.00 kg"#.

More specifically, you will have

#2000 color(red)(cancel(color(black)("g"))) * ("1 mole Fe"_2"O"_3)/(159.7color(red)(cancel(color(black)("g")))) = "12.52 moles Fe"_2"O"_3#

#2000 color(red)(cancel(color(black)("g"))) * "1 mole CO"/(28.01color(red)(cancel(color(black)("g")))) = "71.40 moles CO"#

If all the moles of iron(III) were to take part in the reaction, then the reaction would also consume

#12.52 color(red)(cancel(color(black)("moles Fe"_2"O"_3))) * (color(red)(3)color(white)(a)"moles CO")/(1color(red)(cancel(color(black)("mole Fe"_2"O"_3)))) = "37.56 moles CO"#

The rest of the moles of carbon monoxide will be in excess, i.e. they will not take part in the reaction.

For the last part, you got the approach down correctly. All you have to do here is to double the number of moles of iron(III) oxide and multiply the result by the molar mass of iron metal.

The reaction will produce

#12.52 color(red)(cancel(color(black)("moles Fe"_2"O"_3))) * (color(blue)(2)color(white)(a)"moles Fe")/(1color(red)(cancel(color(black)("mole Fe"_2"O"_3)))) = "25.04 moles Fe"#

This is equivalent to

#25.04 color(red)(cancel(color(black)("moles Fe"))) * "55.845 g"/(1color(red)(cancel(color(black)("mole Fe")))) = color(green)(bar(ul(|color(white)(a/a)color(black)(1.40 * 10^(3)"g")color(white)(a/a)|)))#

The answer must be rounded to three sig figs, the number of sig figs you have for the masses of the two reactants.