Chemistry Stoichiometry Help
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Chemistry Stoichiometry Help

[From: ] [author: ] [Date: 12-03-11] [Hit: ]
0g Fe * (1mole Fe/ 55.85g Fe) * (2mol Fe2O3 / 4 mol Fe) * (159.=78.percent yield = 40.0g / 78.63g*100% = 50.......
I need help with a stoichiometry question for chemistry class.

40.0 g of Iron (III) oxide is formed by the reaction of 55.0 g of iron with an excess oxygen. What is the percent yield? This is a composition reaction.

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first thing to do is write the balances equation of the reaction:
4Fe + 3O2 ---> 2Fe2O3

the question already telling you the limiting reactant is iron and also giving you the actual yield of the eaction.. you need to determine the theoretical yield of the reaction
grams of Fe (use molar mass) --> moles Fe ---> moles Fe2O3 --> grams of Fe2O3 (use molar mass)
55.0g Fe * (1mole Fe/ 55.85g Fe) * (2mol Fe2O3 / 4 mol Fe) * (159.7g Fe2O3 / 1 mol Fe2O3) =
=78.63g Fe2O3

percent yield= actual yield / theoretical yield * 100%

percent yield = 40.0g / 78.63g *100% = 50.9%
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