What Volume of Oxygen Gas is Needed to Combust 3.5 Moles of Propane?

[STEMucator]

Homework Statement

Determine the volume of oxygen gas required to completely combust 3.5mole of propane in a barbecue at standard pressure and 28.0°C.

Homework Equations

$PV = nRT$

The Attempt at a Solution

My first question is, is the work I've done here still correct? http://gyazo.com/3a22ee323ec1edff0ea071255cf698ac

I'm a bit confused here. If i take the standard mole ratio route, I get a completely different answer. I would find that 17.5mole of oxygen would be needed for 3.5mole of propane. Then using the relevant equation I get 437.72925L = 400L because 100kPa only has 1 significant digit. This answer is clearly very far off.

This leads me to believe the question is intending me to do the work as I've shown in the image above.

Why does this incredibly different answer occur? Is it merely a fault of significant error? Or am i wrong completely and the mole route is appropriate (which really wouldn't make sense)?

 

[Borek]

If it is a standard pressure, 100 has at least three significant figures, if not infinitely many. Doesn't matter much, as 3.5 has only two and it is what limits number of digits on the final answer.

I don't see any difference between both approaches. I got 438 L in both cases, rounded down to 440 L. You got different final results as you rounded down intermediate results - never do that. Round down only the final result, use full precision during calculations.

 

[STEMucator]

If it is a standard pressure, 100 has at least three significant figures, if not infinitely many. Doesn't matter much, as 3.5 has only two and it is what limits number of digits on the final answer.

I don't see any difference between both approaches. I got 438 L in both cases, rounded down to 440 L. You got different final results as you rounded down intermediate results - never do that. Round down only the final result, use full precision during calculations.

Ah so I would keep the volume calculation at $88L$. I thought because 100 had only 1 sig fig it would change the answer dramatically.

Doing that I get 440L directly by using the ratio 1/88 = 5/z.

 

[Borek]

No idea where you got 88 L from - unless it is already rounded 87.637. And I just told you to not round down intermediate results, haven't I?

 

[STEMucator]

No idea where you got 88 L from - unless it is already rounded 87.637. And I just told you to not round down intermediate results, haven't I?

Yes you're correct. I did round it.

I have a question now though just for clarity. Is it technically wrong to use significant figures in every step of a problem however many calculations there may be?

 

[Borek]

Yes, it is technically wrong. Round down only the final result, use full precision (or at least several so called guard digits) on all intermediate results. Otherwise you are introducing hard to detect errors (you have just witnessed such a situation).

Note - if you are REPORTING intermediate results, report them rounded.