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

2 Answers
Feb 21, 2018

Answer:

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.

Feb 21, 2018

Answer:

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