# What is the balanced equation of the reaction between gaseous propane and and oxygen gas?

Jun 20, 2018

The typical rigmarole is to....

#### Explanation:

$\text{(i) balance the carbon as carbon dioxide...}$

$\text{(ii) balance the hydrogens as water...}$

$\text{(iii) ...and THEN balance the oxygens......}$

And so we completely COMBUST propane, ${C}_{3} {H}_{8}$..

$\text{(i)}$ ${\underbrace{{C}_{3} {H}_{8} \left(g\right) + {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right)}}_{\text{carbons balanced}}$

$\text{(ii)}$ ${\underbrace{{C}_{3} {H}_{8} \left(g\right) + {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(l\right)}}_{\text{carbons and hydrogens balanced}}$

$\text{(iii)}$ ${\underbrace{{C}_{3} {H}_{8} \left(g\right) + 5 {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(l\right)}}_{\text{the entire equation balanced}}$

Now this works well for odd-numbered alkanes....for EVEN-NUMBERED alkanes...we reach a problem...

${\underbrace{{C}_{4} {H}_{10} \left(g\right) + \frac{13}{2} {O}_{2} \left(g\right)}}_{\text{266 g" rarr underbrace(4CO_2(g) + 5H_2O(l))_"266 g}}$

...OR....

${\underbrace{2 {C}_{4} {H}_{10} \left(g\right) + 13 {O}_{2} \left(g\right)}}_{\text{532 g" rarr underbrace(8CO_2(g) + 10H_2O(l))_"532 g}}$

Sometimes the latter equation is preferred because if you use a half-integral coefficient, you might die. I tend to find the stoichiometry of the FORMER reaction a bit easier to use when calculating stoichiometric equivalence. In either scenario CHARGE and MASS are balanced ABSOLUTELY....as indeed they must be if we purport to represent an actual chemical reaction.

Jun 20, 2018

C_3H_8 + 5O_2 → 3CO_2 +4 H_2O

#### Explanation:

The chemical equation for this reaction is C_3H_8 + O_2 → CO_2 + H_2O, producing heat. To balance the equation, start by balancing $C$,

C_3H_8 + O_2 → 3CO_2 + H_2O

then $H$,

C_3H_8 + O_2 → 3CO_2 +4 H_2O

and finally $O$. Since the product side contains $\left(3\right) \left(2\right) + 4 \left(1\right) = 10$ moles of oxygen, reactant ${O}_{2}$ must be multiplied by 5 to have 10 moles as well.

The balanced chemical equation would be: C_3H_8 + 5O_2 → 3CO_2 +4 H_2O.

HOWEVER,

if too much or too little of ${O}_{2}$ is present, the equations become

2C_3H_8+9O_2→4CO_2+2CO+8H_2O, producing carbon monoxide (soot)

or

C_3H_8+2O_2→3C+4H_2O producing carbon, respectively.