What mass of water would result given complete combustion of a 44.0*g mass of propane?

1 Answer
Jun 5, 2016

Approx. $72 \cdot g$.

Explanation:

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

$\text{Moles of propane}$ $=$ $\frac{44.0 \cdot g}{44.10 \cdot g \cdot m o {l}^{-} 1}$ $\cong$ $1 \cdot m o l$. Given the balanced equation, how many moles of water will result, and how much mass does this constitute?