# A 5.57*g mass of propane was combusted 24.1*g oxygen gas. Can you represent the reaction by means of an equation, and identify the reagent in deficiency?

Apr 26, 2017

So we work out the molar quantities of each reagent.......and interrogate the stoichiometric reaction..........

#### Explanation:

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

$\text{Moles of propane} = \frac{5.57 \cdot g}{44.1 \cdot g \cdot m o {l}^{-} 1} = 0.126 \cdot m o l$

$\text{Moles of dioxygen} = \frac{24.1 \cdot g}{32.0 \cdot g \cdot m o {l}^{-} 1} = 0.753 \cdot m o l$

And thus there is a SLIGHT stoichiometric excess of dioxygen gas. And ${C}_{3} {H}_{8}$ is the reagent in deficiency.

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