Question #73616

1 Answer
Nov 19, 2015



Start with the balanced chemical equation for this reaction

#color(blue)(4)"Fe"_3"O"_text(4(s]) + "O"_text(2(g]) -> color(red)(6)"Fe"_2"O"_text(3(s])#

Now, here's what you're dealing with here. You magnetite ore contains magnetite, #"Fe"_3"O"_4#, and other minerals. When you react this ore with oxygen, a certain mass of ferric oxide, #"Fe"_2"O"_3#, will be formed.

Your strategy here is to figure out how many moles of ferric oxide you have, then use the #color(blue)(4): color(red)(6)# mole ratio that exists between the two compounds to find the number of moles of magnetite in the ore.

So, ferric oxide's molar mass is equal to #"159.69 g/mol"#, which means that every mole of ferric oxide has a mass of #"159.69 g"#. This means that the sample produced by the reaction will contain

#0.5color(red)(cancel(color(black)("g"))) * ("1 mole Fe"_2"O"_2)/(159.69color(red)(cancel(color(black)("g")))) = "0.003131 moles Fe"_2"O"_3#

This means that ore contained

#0.003131color(red)(cancel(color(black)("moles Fe"_2"O"_3))) * (color(blue)(4)" moles Fe"_3"O"_4)/(color(red)(6)color(red)(cancel(color(black)("moles Fe"_3"O"_4)))) = "0.002087 moles Fe"_3"O"_4#

Finally, use magnetite's molar mass, which is equal to #"231.53 g/mol"#, to determine how many grams would contain this many moles

#0.002087color(red)(cancel(color(black)("moles"))) * "231.53 g"/(1color(red)(cancel(color(black)("mole")))) = "0.483 g Fe"_3"O"_4#

This means that the percent composition of magnetite in the ore is

#"% magnetite" = (0.483color(red)(cancel(color(black)("g"))))/(0.88color(red)(cancel(color(black)("g")))) xx 100 = 54.89%#

Now, the answer should be rounded to one sig fig, the number of sig figs you have for the mass of ferric oxide, but I'll leave it rounded to two sig figs, just for good measure

#"% Fe"_3"O"_4 = color(green)(55%)#

SIDE NOTE I don't like using rounded values for the molar masses of various elements. If you want, you can redo the calculations using the molar masses of magnetite and ferric oxide you'd get from

#"Fe"_3"O"_4:" " 3 xx 56 + 4 xx 16 = "232 g/mol"#

#"Fe"_2"O"_3: " " 2 xx 56 + 3 xx 16 = "160 g/mol"#

It's worth mentioning that you should get the exact same result.