# What mass of water results from complete combustion of a 576*g mass of propane?

May 9, 2017

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

#### Explanation:

Is the equation balanced? It must be if it represents physical reality. What does $\Delta$ mean?

And thus each mole of propane results in FOUR moles of water upon complete combustion........

And so we set up the stoichiometric ratio:

$\frac{576 \cdot g}{44.1 \cdot g \cdot m o {l}^{-} 1} \times 18.01 \cdot g \cdot m o {l}^{-} 1 \times 4 = 235 \cdot g$ water........