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
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.
To find the volume of CO2 produced in the reaction, we need to calculate the amount of ethanol used, then determine the stoichiometry of the reaction between ethanol and oxygen gas, and finally, use the percent yield to calculate the actual volume of CO2 produced.
Step 1: Calculate the amount of ethanol used:
Given that the density of ethanol is 0.789 g/ml, and the volume of ethanol used is 50.0 ml, we can calculate the mass of ethanol using the formula:
mass = volume * density = 50.0 ml * 0.789 g/ml = 39.45 g
Step 2: Determine the stoichiometry of the reaction:
The balanced chemical equation for the reaction between ethanol (C2H5OH) and oxygen (O2) to produce carbon dioxide (CO2) and water (H2O) is:
C2H5OH + O2 -> CO2 + H2O
We can see that the stoichiometric ratio for the reactants is 1:1. This means that for every mole of ethanol reacted, one mole of CO2 is produced.
Step 3: Calculate the moles of ethanol:
To determine the moles of ethanol, we need to know the molar mass of ethanol. The molar mass of ethanol (C2H5OH) can be calculated by summing the atomic masses of its constituent atoms:
C: 2 atoms * atomic mass of carbon = 2 * 12.01 g/mol = 24.02 g/mol
H: 6 atoms * atomic mass of hydrogen = 6 * 1.01 g/mol = 6.06 g/mol
O: 1 atom * atomic mass of oxygen = 1 * 16.00 g/mol = 16.00 g/mol
Total molar mass = 24.02 g/mol + 6.06 g/mol + 16.00 g/mol = 46.08 g/mol
Now, we can calculate the moles of ethanol:
moles of ethanol = mass / molar mass = 39.45 g / 46.08 g/mol ≈ 0.855 mol
Step 4: Calculate the moles of CO2 produced:
Since the stoichiometry of the reaction is 1:1, the moles of CO2 produced will be the same as the moles of ethanol reacted:
moles of CO2 = moles of ethanol ≈ 0.855 mol
Step 5: Calculate the volume of CO2 produced:
At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of CO2 produced can be calculated as:
volume of CO2 = moles of CO2 * molar volume at STP = 0.855 mol * 22.4 L/mol ≈ 19.1 L
Finally, the approximate volume of CO2 produced in the reaction is 19.1 liters, given the 78.5% yield.