Question

The combustion of ethanol, C2H5OH produces CO2 and H2O. The density of ethanol is 0.789 g/ml, oxygen is 1.43 g/L, and CO2 is 1.98 g/L. In one experiment 17.6 mL of ethanol were combined with 10.5 L O2 and allowed to react.

Determine the theoretical yield of CO2 in grams.

Answer 1

C2H5OH + 3O2 ==> 2CO2 + 3H2O
mols C2H5OH = 17.6 mL x 0.789 g/mL x (1 mol/46 g) = 0.3018
mols O2 = 10.5 L x 1.43 g/L x (1 mol/32 g) = 0.469
So you need 3 mols O2 for every 1 mol O2. Do you have that? 3*0.3018 = 0.905 g O2 but you don't have that so O2 must be the limiting reagent.
How much CO2 will that produce? That's 0.469 mols O2 x (2 mols CO2/3 mols O2) = 0.313 mols O2 and that x 44 = 13.8 grams CO2 = theoretical yield in grams.

Answer 2

oops! typo. Should be CO2 and not O2 in the last two lines.
That's 0.469 mols O2 x (2 mols CO2/3 mols O2) = 0.313 mols "CO2" and that x 44 = 13.8 grams CO2 = theoretical yield in grams.