Convert 50.0 mL ethanol to mass using mass = volume x density. I obtained approximately 50 x 0.789 = about 40g but that's an estimate only as are all of the other numbers. You need to go through the calculations yourself.
mols ethanol = 40/molar mass = about 0.87
mols O2 = 25.0 L x (1 mol/22.4L) = about 10.41.
C2H5OH + 3O2 ==> 2CO2 + 3H2O
This is a limiting reagent problem.
If we used 0.87 mols ethanol and all of the oxygen needed we could produce 2*0.87 = about 1.7 mols CO2
If we used 10.4 mols O2 and all the ethanol needed we could produce about 10.41 x (2/3) = about 7 mols CO2.
You see the values for mols CO2 produced do not agree which means one is not right. In limiting reagent problems the correct value is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. So ethanol is the limiting reagent and 1.7 mols CO2 are produced.
L CO2 produced = mols CO2 x 22.4 L/mol if it were at 100% yield.
L CO2 actually produced at 78.5% yield will be L at 100% yield x 0.785 = ?L.
You react 25.0 liters of oxygen gas at STP with 50.0 ml of ethanol liquid. Ethanol has a density of 0.789 g/ml.
After actually running the reaction you determined the percent yield was only 78.5%. How many LITERS of CO2 did you make? (Must put CO2 in answer la
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