35.0 g of C 6 H 6 reacts to produce 23.0L of CO 2 at STP. What is the percent yield of the reaction?

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The complete details of the rection are not given. Assuming that you are referring to the reaction of benzene (C6H6) and oxygen (O2) that generates carbon dioxide (CO2), water (H2O), and heat, the balanced chemical reaction can be written as

C6H6 + 15/2 O2 -> 6CO2 + 3H2O + heat.

Here, 1 mole of benzene reacts with 15/2, or 7.5, moles of oxygen to generate 6 moles of carbon dioxide, 3 moles of water, and heat.

Also, note that at STP (standard temperature and pressure), each mole of an ideal gas occupies 22.4 l of volume.

We are given with 35.0 g of benzene, which is equal to 35/78 = 0.45 moles of benzene.

As per the balanced equation, 1 mole of benzene produces 6 moles of carbon dioxide. And hence, 0.45 moles of benzene should theoretically produce 6 x 0.45 moles = 2.7 moles of carbon dioxide.

At STP, 2.7 moles of carbon dioxide will occupy 2.7 x 22.4 l = 60.48 l of volume.

However, as per the given question, only 23.0 l of carbon dioxide has been generated.

Thus, the percent yield of the reaction = 100 x (actual yield/theoretical yield)

= 100 x (23/60.48) = 38%

Therefore, the percent yield of the reaction is 38% only, and hence we can also say that the yield is very low for this reaction.

Hope this helps.

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