Some Chemistry Questions

For some reason, I had the urge to share some Chemistry questions I’ve encountered during my exploits (lol) at Yahoo! Answers. Please study them at your leisure.

Question 1: What volume of carbon dioxide, CO₂, at STP will be produced when 0.250 mol of ethyl alcohol, C₂H₅OH, is burned completely?

The question signifies that a combustion reaction took place. The phrase “burned completely” serves as a context clue. Hence, we can form this balanced chemical equation based on these gathered facts.

C₂H₅OH + 3O₂ ⇒ 2CO₂ + 3H₂O

Let’s now look on what is given. We know that at standard pressure and temperature (STP), pressure is 1 atm and the temperature is 273 K. We’re also given the number of moles of ethanol, which is 0.250 mol. However, these data are insufficient, because we still don’t know the mass of carbon dioxide that was formed, preventing us from using the ideal gas law. Hence, we should apply stoichiometry. Since oxygen gas is in excess, ethyl alcohol becomes the limiting reactant.

My method is unconventional, but I feel comfortable in using this makeshift method, which is also based on the official one. Please bear with me. Let the product take x, and the limiting reactant take y. (? =  coefficient in the balanced equation)

mass of CO₂ = (mole of y) * (? mol x / ? mol y) * (molar mass x)
mass of CO₂ = (0.250 mol C₂H₅OH) * (2 mol CO₂ / 1 mol C₂H₅OH) * (44.01 g/mol CO₂)
mass of CO₂ = 22.0 g

We can now use the ideal gas law to find the volume of carbon dioxide that was formed during the reaction.
PV = nRT
V = nRT / P
V = (22.0 g / 44.01 g/mol)(0.0821 L.atm/mol.K)(273 K) / 1 atm
V = 11. 2 L (Answer)

Question 2: A gaseous sample containing a mixture of only krypton and neon, occupies a volume of 3.10 L at 30 °C. mixture .The partial pressure of Kr is 0.140 atm and that of Ne is 0.470 atm.

What is the total pressure of the mixture?
What is the mole fraction of Kr in the mixture?
How many moles of Kr are in the sample?
What is the total mass of the sample?

Let us not be baffled by just a couple of questions, although I admit that I was a bit daunted at first. First, let us work out the first question, which happens to be exceptionally easy. The total pressure is just the combined partial pressures of the gases involved. In summary, the total pressure is 0.140 atm + 0.470 atm = 0.610 atm
In solving for the mole fraction of Kr in the mixture, which is the ratio between the number of moles of Kr and the total number of moles, we use this formula.

P(a) = X(a)P(t)
P(a) = partial pressure
X(a) = mole fraction
P(t) = total pressure

Since we know the partial pressure of Kr, which is 0.140 atm and the total pressure, which is 0.610 atm, we can solve for the partial pressure quite easily.

0.140 atm = X(a)(0.610 atm)
0.230 = X(a)

The next question being asked is to find the number of moles of Kr present. To do this, we mus first find the total number of moles in the mixture in the first place, using the good ol’ ideal gas law.

PV = nRT
(0.610 atm)(3.10 L) = n(0.0821 L.atm/mol.K)(303 K)
1.891 mol = 24.8763n (Divide both sides by 24.8763)
0.0760 mol = n

We can now find the number of moles of Kr using the formula for mole fraction. Remember that it’s the ratio between the number of moles of a component of a mixture and the total number of moles in the mixture.

X(a) = n(a) / n(t)
0.230 = n(Kr) / 0.0760 mol
0.0175 mol = n(Kr)

Finally, we are asked to find the mass of the whole mixture. Well, since we already the number of moles of Kr present, we can easily find its mass using the fact that mole = mass / molar mass.

0.0175 mol = mass / 83.80 g/mol
1.47 g = mass

We have 1.47 grams of Krypton. To find the mass of Ne, we must first find the number of moles, which, again, is easily done, since the number of moles is additive. So we can say that:

n(Ne) = n(t) – n(Kr)
n(Ne) = 0.0760 mol – 0.0175 mol
n(Ne) = 0.0585 mol

Converting this to mass: 0.0585 mol = mass / 20.18 g/mol ⇒ 1.18 g
Hence, the total number of moles in the mixture is: 1.47 g + 1.18 g = 2.65 g

Hope you enjoyed reading my solution! May this also help other guys struggling at Chemistry. ‘Til next time, folks. Adieu.