# 2C31H64 + 63O2 --> 62CO2 + 64H2O Whatmass of water will be produced in the reaction? Mass before burning 0.72 g after burning 0.27 g

May 6, 2017

We write first the stoichiometric equation.........

#### Explanation:

${C}_{31} {H}_{64} \left(s\right) + 47 {O}_{2} \left(g\right) \rightarrow 31 C {O}_{2} \left(g\right) + 32 {H}_{2} O \left(l\right)$

Now ${C}_{31} {H}_{64}$ is probably a wax or even a tar......Practically, we would expect incomplete combustion with such a hydrocarbon.

If we assume complete combustion, then.............(and I assume, perhaps wrongly, that a $0.5 \cdot g$ mass combusts)

$\text{Moles of hydrocarbon} = \frac{0.50 \cdot g}{436.85 \cdot g \cdot m o {l}^{-} 1} = 1.14 \times {10}^{-} 3 \cdot m o l$.

And the equation CLEARLY indicates that $1.14 \times {10}^{-} 3 \cdot m o l \times 32 = 3.66 \times {10}^{-} 2 \cdot m o l$ water result, i.e. a mass of $0.659 \cdot g$.