Question

How many grams of oxygen are needed to produce .60 liters of carbon dioxide at stp?

Answer 1

Well, of course, we need to define `"STP"`....

Explanation:

...which is hard these days, because the professional bodies have conspired to make the definitions as confusing and as inaccessible as possible.

IUPAC specifies a temperature of `273.15*K`, and a pressure of `0.987*atm`...and under these conditions ONE mole of ideal gas expresses a volume of `22.4*L`...

And so if we gots the combustion reaction....

`C(s) + O_2(g) rarr CO_2(g)`...I need an equiv quantity of dioxygen gas if `0.60*L` carbon dioxide are evolved....(take that atmosphere!)..

And so `"moles of dioxygen gas"-=(0.60*L)/(22.4*L*mol^-1)=0.027*mol`, i.e. a mass of `0.027*molxx32.00*g*mol^-1=0.86*g`.

When you sit an exam, the standard that you use, including the molar volume, should be printed as supplementary material.

Answer 2

You need 0.86 grams of oxygen to produce 0.60 liters of carbon dioxide at STP.

Explanation:

Oxygen has a molar mass of 16.00 g/mol. One mole of CO2 contains 32.00 grams of oxygen.

One mole equals 22.4L at standard temperature and pressure.

0.60L CO2 equals 0.0268 mol at STP.

(32g O) x (0.0268 mol) = 0.8576 g O