Percent Composition of Gas Mixture

1. Nov 18, 2013

Qube

1. The problem statement, all variables and given/known data

It's such a gory problem, I'm gonna have to attach a small image.

http://i.minus.com/j9XQTP6pIlh1a.png [Broken]

2. Relevant equations

PV=nRT

Molarity (M) = moles/liter

3. The attempt at a solution

Okie dokie.

1) Let's find how much iodide reacted with the sulfur dioxide. There appears to be some iodide left over; hence the second reaction. So the iodide that reacted must be the total iodide added minus the iodide that reacted in the second equation.

The moles of iodide added is the volume multiplied by the molarity of the iodide, since the volume units will cancel out. Volume must be converted to liters. All in all, we get that she added 2.034 * 10^-4 moles of iodide.

What part of that is actually used? Well, the amount of excess iodide is similarly [(11.37 mL / 1000) * 0.0105 M thiosulfate]/2, since thiosulfate reacts with the remaining iodide in a 1:2 ratio.

That's 5.969*10^-5 moles of iodide excess.

Therefore the amount of iodide that reacted with SO2 is 2.034 * 10^-4 moles of iodide - 5.969*10^-5 moles of iodide = 1.43 * 10^-4 moles.

2) How many moles of SO2 were there?

Well, SO2 reacts with iodide in a 2:1 ratio. So there must have been twice the number of moles of SO2 as there were iodide that actually reacted. Or in other words, 2.87 * 10^-4 moles.

Okay.

3) V = nRT/P.

We can plug in numbers into the ideal gas law now. T = 311 K. P = 70/76 atm.

I get the volume is 0.00796 L.

This isn't an answer choice, and it's because we're looking for a percentage, not an absolute value.

Since the liters of air is 500 mL or 0.5 L, I divide the volume of SO2 by 0.5, effectively multiplying it, and I get 0.0159. Or 1.59%.

Questions:

1) I know my answer is correct. Is my reasoning sound?

2) Is there a faster way to do this? An alternate way to do this? Or is this pretty much the standard process: find what reacted, how much of what reacted - basically - going backwards?

Last edited by a moderator: May 6, 2017
2. Nov 18, 2013

Staff: Mentor

Yes. When they say volume percent, what they really mean is mole percent. You therefore use the ideal gas law to calculate the number of moles of gas there are, and you know how many moles of SO2 are in the sample. So, you then calculate the mole percent of SO2.

3. Nov 19, 2013

Qube

This is true because volume is proportional to number of moles, correct?

4. Nov 19, 2013

Staff: Mentor

I've always had trouble accepting the concept of assigning part of the total volume to each species in a gaseous solution of 2 or more intimately mixed species. So I never understood the rationale in calling the mole fraction of a species in a gas its volume fraction.

Chet

5. Nov 19, 2013

Staff: Mentor

Technically these are identical, whether the gases are mixed, or separated. (At least as long as the gas is well approximated as an ideal gas).