Partial Pressure Q: Solving for CO2 in Closed Container

[Cyto]

I have this question about partial pressure, and I'm not entirely sure what it is, and how to solve for it...

Q. Solid glucose, C6H12O6 (s) is burned in excess oxygen in a closed container. After the reactiopn, the total gas volume is 10.0L at a temperature of 298K. Wha is the PARTIAL PRESSURE, in atmospheres, of the CO2 (g) produced by the complete combustion of 2.6 grams of glucose?

 

[chem_tr]

Hello,

Partial pressure, as I remember, is calculated as the ratio of one particular gas' pressure to the total. The pressure is directly proportional to the amount, so you can solve this problem by the well-known equation $$P*V=n*R*T$$, where you may further simplify this equation by calculating $$\frac{R*T}{V}$$ as these are constants. You can find how many moles are there in 2,6 grams of glucose as C:12, H:1, and O:16 grams/mol.

Regards,
chem_tr

 

[Bystander]

(snip)Partial pressure, as I remember, is calculated as the ratio of one particular gas' pressure to the total. (snip)

Gotta correct this --- you've stated one of several ways to calculate "mole fraction."

Without getting into dictionary circles (partial pressure, mole fraction times total pressure, etc.), "partial pressure" is the pressure exerted by whatever species happens to be of interest in the absence (hypothetical) of all other gas species in the system.

 

[chem_tr]

Yes, you are right. We answer the hypothetical question, "what would happen if there were only one type of gas in the same volume?" by using partial pressures. Thank you, bystander, for correcting me.