# How many moles of water can be produced from the reaction of 28 g of C_3H_8?

May 24, 2017

I presume you speak of a combustion reaction? All hydrocarbons combust completely to give carbon dioxide and water.........Here we gets approx. $2.5 \cdot m o l$.

#### Explanation:

First we need a stoichiometric equation, and this is something at which you need to be very adept. The usual rigmarole is: balance the $\text{carbons (as carbon dioxide)}$; then balance the $\text{hydrogens (as water)}$; then balance the $\text{oxygens}$.

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

$\text{Balance the carbons..........}$

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

$\text{Balance the hydrogens..........}$

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

$\text{Balance the oxygens..........}$

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

With an even-numbered alkane, we would need a half-integral coefficient; try it out for butane, ${C}_{4} {H}_{10}$.

${C}_{3} {H}_{8} \left(g\right) + 5 {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(l\right)$
We know that $1 \cdot m o l$ propane gives $3 \cdot m o l$ carbon dioxide, and $4 \cdot m o l$ of water......In this reaction, we combusted a molar quantity of...................
$\text{Moles of propane} = \frac{28.0 \cdot g}{44.1 \cdot g \cdot m o {l}^{-} 1} = 0.635 \cdot m o l$.
And given the stoichiometry of the reaction, the water that arises from the combustion is 0.635*molxx4=??