# A 30*L volume of dioxygen gas at 473*K, and a pressure of 2*atm, will completely combust WHAT mass of "ethane gas"?

Nov 14, 2015

We need to work out (i) moles of oxygen gas, and (ii) equivalent moles of ethane.

#### Explanation:

$n = \frac{P V}{R T}$ $=$ $\frac{\left(2 \cdot a t m\right) \left(30.0 \cdot L\right)}{\left(0.0821 \cdot L \cdot a t m \cdot {K}^{- 1} \cdot m o {l}^{-} 1\right) \left(473 \cdot K\right)}$

$=$ $1.54$ $m o l$ ${O}_{2} \left(g\right)$.

Now, combustion reaction of ethane can be given as:
${C}_{2} {H}_{6} \left(g\right) + \frac{7}{2} {O}_{2} \left(g\right) \rightarrow 2 C {O}_{2} \left(g\right) + 3 {H}_{2} O \left(g\right)$

So, this $1.54$ $m o l$ quantity represents 7/2 equiv of ethane by the stoichiometry of the rxn. Thus moles of ethane = $1.54 \cdot m o l \times \frac{2}{7}$ $=$ $0.440$ $m o l$ ethane.

So (finally!), mass of ethane $=$ $0.440 \cdot m o l \times 30.07 \cdot g \cdot m o {l}^{- 1}$ $=$ ??g. Approx. 12 g?