M 15

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Jarrett Peyrefitte 2K
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M 15

Postby Jarrett Peyrefitte 2K » Thu Oct 03, 2019 12:19 am

I'm having trouble with letter c in particular. What steps should I take to solve the percentage yield of aluminum chloride?

"Aluminum metal reacts with chlorine gas to produce aluminum chloride. In one preparation, 255 g of aluminum is placed in a container holding 535 g of chlorine gas. After reaction ceases, it is found that 300. g of aluminum chloride has been produced. (a) Write the balanced equation for the reaction. (b) What mass of aluminum chloride can be produced by these reactants? (c) What is the percentage yield of aluminum chloride?"

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Re: M 15

Postby Chem_Mod » Thu Oct 03, 2019 12:21 am

First, identify the limiting reactant based off the starting amounts of each reactant. Then, calculate the maximum theoretical yield of the product based off the amount of the reactant that you have identified as limiting. Finally, use the actual yield given in the problem and the formula for percent yield

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Re: M 15

Postby JustinHorriat_4f » Thu Oct 03, 2019 12:29 am

Yes so the process is
1) find the limiting reactant
2) find the theoretical yield
3) use the formula of percent yield which is actual/theoretical x 100%
it is very simple just make sure you have your numbers correctly, and I wouldn't round to sig figs during the process of calculations because u might get a wrong percentage by leaving numbers out. Also, in a problem like this, you are usually always given the actual yield.

Junwei Sun 4I
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Re: M 15

Postby Junwei Sun 4I » Thu Oct 03, 2019 12:33 am

Firstly you will need to convert grams to moles and check which reactant is the limiting reactant. Once you know the limiting reactant you should use the molar ratio and the balanced chemical equation to calculate the amount of aluminum chloride you will get(this is the theoretical yield.) Then use the formula percent yield = (actual yield/theoretical yield)*100% to calculate the final answer to part (c). BTW actual yield is given in the question.

Indy Bui 1l
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Re: M 15

Postby Indy Bui 1l » Thu Oct 03, 2019 12:46 am

You need to solve parts a) and b) in order to solve part c. The mass of AlCl2 you calculated in part B) is used as the theoretical yield. The question gives the actual yield as 300g. All you need to do is solve the equation 100% * (Actual Yield)/(Theoretical Yield). For my answer I got 100%* ((300g)/(671g))=44.7%

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