I have no idea what the reaction is.
I am surprised your chem teacher has not counseled you yet on the math issue.
What is the theoretical yield of H[2]O (in grams) if 8.3 mol H[2]S were consumed in the above reaction? If 137.1g of water was actually collected after the reaction, what is the percent yield?
I have no idea to do this problem. I am not good in math at all and currently taking algebra 1 which I did not pass first semester so you can see that chemistry is killing me now. Thanks. Ty
5 answers
I have an IEP and they don't care and they aren't following the IEP guidelines either. I am drowning.
Did you read Bob Pursley's message? He is telling you that you didn't write the "above reaction". We can't read your mind and you wrote the problem part but not all of the question. Write the equation and we may be able to help.
2H[2}S(g) + 30[2] (g) arrow 2SO[2] (g) + 2H[2]O (g)
8.3 mol H2S used. Use the coefficients in the balanced equation to convert mols H2S to mols H2O.
That's 8.3 mol H2S x (2 mols H2O/2 mols H2S) = 8.3 x 2/2 = 8.3 mols H2O formed.
Convert that to grams.
grams H2O = mols H2O x molar mass H2O = estimated 149 g (but you need to do it more accurately) and that is the theoretical yield. The problem gives the actual yield of 137.1 g H2O.
% yield = (actual yield/theor yield)*100 = (137.1/149)*100 = ? Remember to confirm my estimates above and use YOUR numbers in the percent yield calculation and mass H2O produced.
That's 8.3 mol H2S x (2 mols H2O/2 mols H2S) = 8.3 x 2/2 = 8.3 mols H2O formed.
Convert that to grams.
grams H2O = mols H2O x molar mass H2O = estimated 149 g (but you need to do it more accurately) and that is the theoretical yield. The problem gives the actual yield of 137.1 g H2O.
% yield = (actual yield/theor yield)*100 = (137.1/149)*100 = ? Remember to confirm my estimates above and use YOUR numbers in the percent yield calculation and mass H2O produced.