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

Feb 21, 2018

Well, of course, we need to define $\text{STP}$....

#### 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 $273.15 \cdot K$, and a pressure of $0.987 \cdot a t m$...and under these conditions ONE mole of ideal gas expresses a volume of $22.4 \cdot L$...

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

$C \left(s\right) + {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right)$...I need an equiv quantity of dioxygen gas if $0.60 \cdot L$ carbon dioxide are evolved....(take that atmosphere!)..

And so $\text{moles of dioxygen gas} \equiv \frac{0.60 \cdot L}{22.4 \cdot L \cdot m o {l}^{-} 1} = 0.027 \cdot m o l$, i.e. a mass of $0.027 \cdot m o l \times 32.00 \cdot g \cdot m o {l}^{-} 1 = 0.86 \cdot g$.

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

Feb 21, 2018

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