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Report (any type)/Brief report
22/09/2021
##”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”)##
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I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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