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

Question

"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.

Peyton Adams · Accepted Answer

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))# On this one, I'm not really sure if I understand your calculations. First off, comparing the molar masses of the two reactants can help you quickly determine which one will serve as the limiting reagent in a situation where the masses of the two reactants are equal. When the masses of the two reactants are equal, the reactant with the larger molar mass will contain the fewest moles 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"#. To be more precise, 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"# The reaction would likewise consume all of the moles of iron(III) if they were to participate. #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 remaining moles of carbon monoxide won't participate in the reaction since they will be in excess. You have the methodology down for the last part, which is just multiplying the result by the molar mass of iron metal and doubling the amount of moles of iron(III) oxide. The outcome of the reaction will be #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 the same as #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 number of sig figs you have for the masses of the two reactants, which is three, must be rounded off in the answer.

Owen Ambrose · Answer

undefined Divide the mass of the limiting reactant by its molar mass, then use the balanced chemical equation to find the molar ratio with the desired product. Multiply by the molar mass of the product to determine the mass of the product formed.