# Is a 3.44*g mass of oxygen gas sufficient to completely oxidize a 14.8*g mass of propane?

Sep 4, 2017

${C}_{3} {H}_{8} + 5 {O}_{2} \rightarrow 3 C {O}_{2} + 4 {H}_{2} O$

There are insufficient dioxygen for complete combustion......

#### Explanation:

And then it is simply a matter of working out the molar quantities:

$\text{Moles of propane} = \frac{14.8 \cdot g}{44.10 \cdot g \cdot m o {l}^{-} 1} = 0.336 \cdot m o l$

$\text{Moles of dioxygen} = \frac{3.44 \cdot g}{32.00 \cdot g \cdot m o {l}^{-} 1} = 0.108 \cdot m o l$

Oxygen is present in stoichiometric deficiency. We need further details of the experiment.