If 16g of aluminum oxide were produced, what is the percent yield of the reaction below?
this is the reaction:
4Al+3O2--->2Al2O3
Answer:
4 moles of Al would yield 2 moles of Al2O3
10 g Al is 10/26.98 = 0.3706 moles
You would expect 0.3706/2 moles ( 0.1853 moles) of Al2O3 at 100% yield.
You obtained 16g/101.96 = 0.1569 moles of Al2O3
Your percent yield was 0.1569/0.1853 = 84.7%
Okay, here's an overview of what you need to do:
1) Determine which one of your starting materials is the limiting reagent by comparing molar quantities
2) Determine based on the limiting reagent and assuming 100% yield, the theoretical yield for the reaction
3) Determine percent yield for the reaction by dividing the actual yield (16g) by the theortical yield, then times-ing by 100 to get a percent.
Now here's the specifics:
Molar amounts can be determined by multiplying the initial amount of material by the reciprocal of the molecular weight.
10g Al * (1 mole Al/26.982g Al) = 0.371 moles Al
19g O2 * (1 mole O2/32g O2) = 0.594 moles O2
meaning that the aluminum is the limiting reagent.
Next, for theoretical yield, we need to get the initial grams of limiting reagent to grams of final product (just some stoiciometry):
10g Al * (1mole Al/26.982g Al) * (2mole Al2O3/4moles Al) * (101.964g Al2O3/1mole Al2O3) = 18.895 grams theoretically.
*where the terms (2mole/4mole) is the molar ratio of reagent to product taken from the balanced equation.
Now, all you need is to calculate percent yeild:
(16 grams product actual/18.895 grams theoretically) * 100= 84.67%
Of course, if you're watching your sig figs (as you should!!) the answer simplifies to 85%. Hope that helps!
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