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 label)

To calculate the number of liters of CO2 produced in the reaction, we need to use the given information and the concept of stoichiometry.

First, let's determine the number of moles of ethanol used. We can use the density of ethanol to convert the given volume to mass:

Mass of ethanol = volume of ethanol x density of ethanol
= 50.0 ml x 0.789 g/ml
= 39.45 g

Next, let's convert the mass of ethanol to moles. We can do this by dividing the mass by the molar mass of ethanol, which is 46.07 g/mol:

Moles of ethanol = mass of ethanol / molar mass of ethanol
= 39.45 g / 46.07 g/mol
= 0.855 mol

From the balanced chemical equation, we know that the reaction between ethanol and oxygen produces 2 moles of CO2 for every 1 mole of ethanol. Therefore, the moles of CO2 produced can be calculated as:

Moles of CO2 = 2 x moles of ethanol
= 2 x 0.855 mol
= 1.71 mol

Now, we can use the ideal gas law to convert the number of moles of CO2 to volume at STP. The molar volume of a gas at STP is 22.4 L/mol:

Volume of CO2 = moles of CO2 x molar volume
= 1.71 mol x 22.4 L/mol
= 38.344 L

Finally, to account for the 78.5% yield, we multiply the calculated volume by the percent yield expressed as a decimal:

Actual volume of CO2 = volume of CO2 x percent yield (as a decimal)
= 38.344 L x 0.785
= 30.106 L

Therefore, you made approximately 30.106 liters of CO2 in the reaction.