Stoichiometry, Part II (and Percentage Yield)

1) In an oxidation reaction, ethanol (C2H5OH) is converted to a ethanoic acid (C2H4O2) with a percentage yield of 70%. If you start the reaction with 360g of ethanol, how many grams of ethanoic acid would you expect to produce?
Stoichiometry is 1:1 between the compounds.

% yield = (actual value/expected value) *100

Given: % yield = 70

Find: expected value C2H4O2 in g

Actual value C2H4O2 in g:
360 g C2H5OH (1 mol C2H5OH / 46.068 g C2H5OH) (1 mol C2H4O2 / 1 mol C2H5OH) = 7.81454 mol C2H4O2

7.81454 mol C2H4O2 (60.052 g C2H4O2 / 1 mol C2H4O2) = 469.2785 g C2H4O2 = actual value ??

70 = (469.2785/expected) * 100
expected = 469.2785/.70 = 670 g C2H4O2 ??

Did I muddle this problem big time? I'm confused with what is the actual value and what is the expected (theoretical?) value. Did I confuse the actual for the expected?? Do you use 70% at all?

2) In an oxidation reaction, ethanol (C2H5OH) is converted to a ethanoic acid (C2H4O2). In an Experiment, a chemist starts with 149.00g of ethanol and forms 180.00g of ethanoic acid. What is the percentage yield of the reaction? Stoichiometry of the reaction is 1:1 for the mentioned compounds.

% yield = (actual value/expected value) *100

Given: 180.00 g C2H4O2 = actual value g C2H4O2 ?

Expected value of C2H4O2 :
149.00 g C2H5OH (1 mol C2H5OH / 46.068 g C2H5OH) (1 mol C2H4O2 / 1 mol C2H5OH) (60.052 g C2H4O2 / 1 mol C2H4O2) = 194.2291 g C2H4O2 ??

180 g / 194.2291 g C2H4O2 *(100) = 92.674 % ??

Gosh, I hope I didn't confuse my values or molar masses again!

Thank you. I am desperate for help. Anyone at all? Please? (Are my calculations correct?)

Thanks.

Please confirm if I did something correctly.

Your calculations look correct to me. GCT