Someone please help me

Question

Consider the reaction.

2Pb(s)+O2(g)⟶2PbO(s)

An excess of oxygen reacts with 451.4 g of lead, forming  of lead(II) oxide. Calculate the percent yield of the reaction.

percent yield:__________%

Answer 1

the equation:

2Pb + O2 >>>> 2PbO

You are asked to find the percent yield.

% Yield = Actual (experimental) Yield/Theoretical Yield x100.

To calculate the % yield you have either been given the actual yield of this reaction or actually obtained the actual yield through a lab procedure, So the actual yield is an experimental yield and will almost always be less than the theoretical yield that you obtain through stoichiometric calculations from the balanced equation given above. The actual yield is almost always less than the theoretical yield because of spillage of the product, contamination of the products, product loss in the purification process, a decrease in the products because of competing chemical reactions, and product loss because of the reaction not going to completion.

With the information given above in your question you certainly can find the the theoretical yield of the reaction by converting the grams of lead given to moles of lead and then using molar ratios of lead to PbO which by inspection is 1:1 calculate the theoretical yield of PbO in moles. Then taking the moles found of PbO and converting it back to grams.

Unfortunately not knowing the actual (experimental) yield you would not be able to finish the above equation to determine the % yield.

% Yield = Actual Yield (which we don't have presently)/ Theoretical Yield (which we can calculate as mentioned above)

If I might suggest, look carefully for the data given or the data that you found on experiment that would actually give you the value of the Actual Yield.