anonymous
  • anonymous
13.4 grams of iron (III) oxide reacts with 12.3 grams of carbon monoxide to produce 8.23 g of pure iron, what are the theoretical yield and percent yield of this reaction? Be sure to show the work that you did to solve this problem.Please Help
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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aaronq
  • aaronq
1. write a reaction then balance it. 2. find the moles of iron produced by using the stoichiometric coefficients. e.g. for a general reaction: aA + bB \(\rightleftharpoons\) cC where upper case are the species, and lower case are the coefficients, \(\dfrac{n_A}{a}=\dfrac{n_B}{b}=\dfrac{n_C}{c}\) from here, you can solve for moles then convert to grams. The amount of grams you find here is the theoretical yield. 3. finding percent yield, use: \(\% yield=\dfrac{actual\;yield}{theoretical\;yield}*100\%\) the units should both be the same, otherwise it won't work.
anonymous
  • anonymous
Fe2O3 + 3 CO-->2 Fe + 3 CO2
anonymous
  • anonymous
right?

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aaronq
  • aaronq
yup, you now need to convert the masses given to moles and use their stoichiometric coefficients to build a ratio, like i stated above.
anonymous
  • anonymous
can u help with that please
aaronq
  • aaronq
convert to moles first
anonymous
  • anonymous
idk how
aaronq
  • aaronq
13.4 grams of iron (III) oxide reacts with 12.3 grams of carbon monoxide \(n=\dfrac{m}{M}\) where, n=moles, m=mass (in g), M=molar mass (in g/mol). You'll find M on the periodic table
anonymous
  • anonymous
...
aaronq
  • aaronq
...
aaronq
  • aaronq
?
anonymous
  • anonymous
can you just do the question out so i can find the percent yield
aaronq
  • aaronq
i'm not going to do your work
anonymous
  • anonymous
I just need explanation, like a bit of a back and forth thing.
aaronq
  • aaronq
i wrote above how you can find the amount of moles of both reactants.
anonymous
  • anonymous
or can u add someone else to help me please?
anonymous
  • anonymous
ill give a medal
aaronq
  • aaronq
i'm trying to help you but you don't wanna do any work.
anonymous
  • anonymous
@Hero please help
anonymous
  • anonymous
@Chemistry please help

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