Question

15O2(g) + 2C6H5COOH(aq) → 14CO2(g) + 6H2O(g) If 45.0 moles of O2(g) and 10.6 moles of C6H5COOH(aq) are allowed to react, and the percent yield is 91.6%, how many moles of CO2(g) will actually be produced? I know what to do but I keep getting an answer that's not listed. Please help this is due at midnight!

39.5 moles

34.5 moles

11.3 moles

69.5 moles

38.5 moles

Answer 1

15 mole of O2 requires 2 mole of C6H5COOH so

45.0 mole of O2 will require

= 45.0 mole of O2 * (2 mole of C6H5COOH / 15 mole of O2)

= 6 mole of C6H5COOH

but we have 10.6 mole of C6H5COOH which is in excess so O2 is limiting reactant

fromthe balanced equation we can say that

15 mole of O2 produces 14 mole of CO2 then

45.0 mole of O2 will produce

= 45.0 mole of O2 *(14 mole of CO2 / 15 mole of O2)

= 42 mole of CO2

Therefore, theoretical yield of CO2 = 42 mole

% yield = (actual yield/theoretical yield)*100

91.6 = (actual yield / 42) * 100

acual yield = 38.5 mole

Therefore, the actual yield of CO2 is 38.5 mole