Ethanol will combust according to the following equation:
C2H5OH(l)+3O2(g)->2CO2(g)+3H2O(l)
How many liters of air are required to combust 120.1 g of ethanol at 26.4C and 789.5 mmHg. Assume air is 21% O2 by volume.
I cannot find my error - professor states answer as 882.6 +/- .5
Moles of C2H5OH = 120.1 g / 46.069 g/mol = 2.60696 mol C2H5OH.
3x 2.60696 = 7.82088 mol O2
P=789.5 mmHg / 760 mmHg = 1.03882 atm
T=26.4C+273=299.4 K
PV=nRT => V = nRT/P
(7.82088 * 0.08206 * 299.4)/ 1.03882 = 880.804 - Where did I go wrong?
I don't see any correction to go from pure O2 to air (21% O2 in air).
A minor point.
Since T is given as 26.4 I would have added that to 273.2.
But if 882.6 is the correct answer I don't know how you came that close when not using the 21%. Perhaps you just didn't show that step.
(7.82088 * 0.08206 * 299.4)/ 1.03882 = 184.969
184.969 / 0.21 = 880.804
You are correct, I calculated it but forgot to write it in.
To find the volume of air required to combust 120.1 g of ethanol, you made a mistake in the calculation.
The balanced equation tells us that 1 mol of C2H5OH reacts with 3 mol of O2. Therefore, 2.60696 mol of C2H5OH will react with 3 * 2.60696 = 7.82088 mol of O2.
The mole fraction of O2 in air is given as 21%. This means that for every 100 L of air, 21 L is O2. Therefore, the volume of air required to provide 7.82088 mol of O2 is (100 L * 7.82088 mol O2) / (0.21 mol O2/L) = 3724.234 L.
However, you need to convert the conditions to the ideal gas law units before calculating the volume of air.
P = 789.5 mmHg = 789.5 mmHg * 1 atm / 760 mmHg = 1.03882 atm
T = 26.4°C + 273.15 = 299.55 K
Using the ideal gas law equation PV = nRT and rearranging to solve for V, we have:
V = (nRT) / P
V = (7.82088 mol * 0.08206 L·atm/(mol·K) * 299.55 K) / 1.03882 atm
V = 181.409 L
So, the correct volume of air required to combust 120.1 g of ethanol is approximately 181.409 L, not 880.804 L. It appears that your professor's answer of 882.6 +/- 0.5 L may be incorrect, or there could be another mistake in your calculation.
To find the volume of air required to combust 120.1 g of ethanol, you correctly determined the moles of ethanol: 2.60696 mol C2H5OH.
Then, you determined the moles of O2 required for combustion by multiplying the moles of ethanol by the stoichiometric coefficient in the balanced equation: 3 * 2.60696 = 7.82088 mol O2.
To calculate the volume of air required, you used the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
You then rearranged the equation to solve for V: V = nRT/P. You plugged in the values: n = 7.82088 mol, R = 0.08206 L·atm/(mol·K), T = 299.4 K, and P = 1.03882 atm, and calculated V to be 880.804 L.
So far, you have correctly followed the steps. Your calculation matches closely with the expected answer of 882.6 L. However, it appears that you have made a rounding error while reporting the answer. The answer should be 880.8 L, rounded to one decimal place and in agreement with the professor's answer range of 882.6 +/- 0.5 L.