# How do you balance the equation C4H10+O2 = CO2+H2O?

Apr 7, 2018

$\textcolor{b l u e}{2 {C}_{4} {H}_{10} \left(g\right) + 13 {O}_{2} \left(g\right) \to 8 C {O}_{2} \left(g\right) + 10 {H}_{2} O \left(g\right)}$

#### Explanation:

We have the unbalanced equation:

${C}_{4} {H}_{10} \left(g\right) + {O}_{2} \left(g\right) \to C {O}_{2} \left(g\right) + {H}_{2} O \left(g\right)$

This is the combustion of butane, ${C}_{4} {H}_{10}$.

Let's first balance the carbons. There are four on the $\text{LHS}$, but only one on the $\text{RHS}$, so we multiply $C {O}_{2}$ by $4$ to get:

${C}_{4} {H}_{10} \left(g\right) + {O}_{2} \left(g\right) \to 4 C {O}_{2} \left(g\right) + {H}_{2} O \left(g\right)$

Now, let's balance the hydrogens. There are $10$ on the left side, but only two on the right side, so we multiply ${H}_{2} O$ by $5$, and get:

${C}_{4} {H}_{10} \left(g\right) + {O}_{2} \left(g\right) \to 4 C {O}_{2} \left(g\right) + 5 {H}_{2} O \left(g\right)$

Final step is to balance the oxygens. There are two on the left hand side, but thirteen on the right hand side, so we need to divide thirteen by two to get the "scale number", which is $6.5$. The equation is thus:

${C}_{4} {H}_{10} \left(g\right) + 6.5 {O}_{2} \left(g\right) \to 4 C {O}_{2} \left(g\right) + 5 {H}_{2} O \left(g\right)$

But wait, we cannot have half a molecule! So, we need to multiply the whole equation by $2$, which leads us to the finalized, balanced equation:

color(blue)(barul(|2C_4H_10(g)+13O_2(g)->8CO_2(g)+10H_2O(g)|)

Apr 7, 2018

color(teal)(2C_4H_10+13O_2 -> 8CO_2+10H_2O

#### Explanation:

The given equation to be balanced is ${C}_{4} {H}_{10} + {O}_{2} \to C {O}_{2} + {H}_{2} O$

We begin by counting the number of $C$ atoms on both sides.
We have $4 C$ on the left and $1 C$ on the right.

$\implies {C}_{4} {H}_{10} + {O}_{2} \to 4 C {O}_{2} + {H}_{2} O$

color(white)(wwwwwwwwwwww

Now count the number of $H$ atoms on both sides.
We have $10 H$ on the left and $2 H$ on the right.

$\implies {C}_{4} {H}_{10} + {O}_{2} \to 4 C {O}_{2} + 5 {H}_{2} O$

color(white)(wwwwwwwwwwww

Now, $O$'s turn.
We have $2 O$ on the left and $13 O$ on the right.

$\implies {C}_{4} {H}_{10} + \frac{13}{2} {O}_{2} \to 4 C {O}_{2} + 5 {H}_{2} O$

color(white)(wwwwwwwwwwww $\text{or}$

color(teal)(2C_4H_10+13O_2 -> 8CO_2+10H_2O

Lets verify, by counting the number of each atom on each side.

color(white)(wwwwwww $\text{on left }$ color(white)(wwwwwww $\text{on right }$

$C$ color(white)(wwwwwww $8$ color(white)(wwwwwwwwtwww $8$

$H$ color(white)(wwwtwww $20$ color(white)(wwwwwwwwwww $20$

$O$ color(white)(wwwuwww $26$ color(white)(wwwwwwwwwww $26$

Hence the equation is balanced :)