Question

2Pb(s)+O2(g)--->2PbO(s)
In the following reaction, 451.4 g of lead reacts with excess oxygen forming 349.7 g of lead(II) oxide. Calculate the percent yield of the reaction.

Answer 1

find the theoretical yield then use:

\(percent ~yield=\dfrac{theoretical-actual}{theoretical}*100\%\)

Answer 2

which is which? :/ I'm sorry I'm not good at this

Answer 3

actual is what is given in the question.

theoretical is the most you could make - found through stoichiometry

Answer 4

oh okay how do you find the most you could make?

Answer 5

Convert the mass of reactant to moles

\(\sf moles=\dfrac{mass}{Molar~mass}\)

Next, use the stoichiometric coefficients and the moles to find moles of product

\(\sf \dfrac{moles~of~Pb}{Pb's~coefficient}=\dfrac{moles~of~PbO}{PbO's~coefficient}\)

Convert moles to mass (using the first formula)

use the perfect yield formula

Answer 6

@aaronq If actual yield is not mentioned, is there any way to find percent yield?

Answer 7

For instance imagine if we have only theoretical mass and not actual mass and the question has asked us to find the percent yield?

Answer 8

@aaronq Let me know.

Answer 9

@lindsh Do you know the actual yield inside your question?

Answer 10

The actual yield is supposedly given in the problem but I'm not sure. that's why I'm here lol

Answer 11

Yes that is confusing because I am doing the same stuff. The actual yield must be give to find percent yield. Other than that it is almost impossible.

Answer 12

Do not close the question. Let him come back and answer my question. Thx

Answer 13

alright @aaronq I think I got the wrong answer..

Answer 14

if the actual yield is not given there is not way that you could find that out, so no you can't do the question.

Answer 15

@korosh23