How many liters of oxygen are needed to exactly react with 25.8 g of methane at STP? CH_4(g) + 2O_2(g) -> CO_2(g) + 2H_2O(l)?

May 20, 2018

Well at $\text{STP}$ one mole of gas occupies $22.4 \cdot L$ (this depends on syllabus; luckily, the definition is included as supplementary material on most exam papers....)

Explanation:

And so we gots the combustion equation...

$C {H}_{4} \left(g\right) + 2 {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right) + \Delta$

And so $\text{moles of methane} = \frac{25.8 \cdot g}{16.04 \cdot g \cdot m o {l}^{-} 1} = 1.536 \cdot m o l$..

And so we require TWICE this molar quantity of dioxygen gas...

And this constitutes a VOLUME of....

$2 \times 1.536 \cdot m o l \times 22.4 \cdot L \cdot m o {l}^{-} 1 \cong 70 \cdot L$

And why should we study this reaction? Well, it heats our homes, and cooks our dinner, and sometimes drives our cabs. In short it allows us to live beyond 30...