When 84.8 g of iron(III) oxide reacts with an excess of carbon monoxide, iron is produced.

Fe2O3(s) + 3CO(g) ==> 2Fe(s) + 3CO2(g)

what is the theoretical yield of iron?

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  1. Most of these problems, where only one reactant is given, is a simple stoichiometry problem. Print out these instructions and memorize them.
    1. Write the equation and balance it.
    2. Convert grams of iron oxide to moles. moles = grams/molar mass
    3. Using the coefficients in the balanced equation, convert moles iron(III) oxide to moles Fe.
    4. Now convert moles Fe to grams Fe. grams = moles x molar mass. This is the theoretical yield.

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  2. Use the balanced equation to determine the theoretical yield. Focus on the chemicals whose quantities you are given or are asked to find. CO is not important as it is excess and will make sure that all of the Fe2O3 will be changed to iron. Remember the quotients (numbers in front of formula) are the number of moles. You are given the mass of Fe2O3 in grams so you will need to convert 1 mole of Fe2O3 into grams (molar mass which, using simple atomic masses = 56x2 + 16x3 = 160g/mol but you use atomic masses that your course requires

    Fe2O3(s) + 3CO(g)--> 2Fe(s) + 3CO2(g)
    1 mol 2 mol
    160g 112g
    84.8g (112/160) x 84.8

    So now you have the theoretical yield of Fe2O3 and the experimental (actual) yield = 54.3g)

    Percentage Yield = (actual yield/theoretical yield) x 100

    Hope this helps :)

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  3. how did u get the molar ratio of 1 to 2. I was confused and thought it was 3 to 1. (1 iron oxide(|||) and 3 carbon monoxide )

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  4. 92% wc

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  5. sdewf

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