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Question

SO2 + O2 → SO3
If the percent yield of the above reaction is found to be 92.80 %, how many grams of SO3 are produced from the complete reaction of 160.00 g of O2 ?

Answer 1

I worked this up, but Jess beat me to it.

Follow her very clear explanation...excellent presentation.
But correct her math...
10 mols x 0.928 = 9.28 mols SO3 produced
MW of SO3 is 80.06
So, 9.28 x 80.06 = 743 g SO3 produced

Don't be harsh...it's easy to transpose digits. :-))

Answer 2

First balance the equation, which is relativly straightforward.

2 SO2 + O2 ---> 2 SO3

work out the moles of oxygen.

Moles O2 = mass / molecular mass = 160g / 32gmol
= 5 moles of O2

Oxygen is the limiting reagent.

I worked out the Theoretical yield first.
= (5 moles O2)* (2mol SO3/1mol O2)* (80g/1mol SO3)
= 800g

That is the amount produced assuming the reaction went to 100% completion.

Given the equation for % yield = Actual Yield / Theoretical Yield * 100

92.80 = Acutal Yield / 800g * 100

Rearranged this give:

Actual yield = (92.80/100) * 800
= 742.4g SO3

Answer 3

Write a balanced equation first:
2 SO2 + O2 ----> 2SO3

Change grams of O2 to mols of O2
160.00 g O2 * 1 mol O2/ 32.00 g = 5 mol O2

Using the equation, we know that for every one mol of O2 reacted, we get 2 mol of SO3. So if we have 5 mol O2, we'll get
5 mol O2 * 2 mol SO3/ 1 mol O2 = 10 mol SO3

But the rxn only goes to 92.80 % completion
so you'll get 10 (.982) = 9.82 mol of SO3

Convert mol of SO3 to grams
9.82 mol SO3 * 70 g/ mol =687.4 g SO3

Answer 4

2SO2 + O2 => 2SO3
32 160
160 ?
the grams of SO3 produced with 100% 160 * 160 / 32 = 800g
the grams of SO3 with 92.8% 800 * 92.8% = 742.4g

Answer 5

I think 137.888888888888(inf.)