# A 17.8*L volume of propane gas for which rho=493*g*L^-1 was completely combusted. What are (i) the molar quantities, and (ii) the mass of carbon dioxide emitted?

Dec 19, 2016

Approx. $26.4 \cdot k g \cdot C {O}_{2}$ are emitted.

#### Explanation:

We need (i) a stoichiometrically balanced equation:

${C}_{3} {H}_{8} \left(g\right) + 5 {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(l\right)$

And (ii) the equivalent quantity of propane gas.

$\text{Moles of propane}$ $=$ $\frac{17.8 \cdot \cancel{L} \times 493 \cdot \cancel{g} \cdot \cancel{{L}^{-} 1}}{44.1 \cdot \cancel{g} \cdot m o {l}^{-} 1} = 199 \cdot m o l$

Now clearly, almost $600 \cdot m o l$ of carbon dioxide will be evolved. Why clearly?

And thus mass of $\text{carbon dioxide}$ emitted,

$600 \cdot m o l \times 44.01 \cdot g \cdot m o {l}^{-} 1 \cong 9 \cdot k g \cdot C {O}_{2}$