# How do you balance these equations?

## ${H}_{3} P {O}_{4} \left(a q\right) + M g {\left(O H\right)}_{2} \left(a q\right) \to M {g}_{3} {\left(P {O}_{4}\right)}_{2} \left(a q\right) + {H}_{2} O \left(l\right)$ ${C}_{4} {H}_{10} \left(g\right) + {O}_{2} \left(g\right) \to C {O}_{2} \left(g\right) + {H}_{2} O \left(g\right)$

Dec 3, 2016

The same way you balance all equations: conserve mass and charge.

#### Explanation:

${H}_{3} P {O}_{4} \left(a q\right) + M g {\left(O H\right)}_{2} \left(a q\right) \rightarrow M g H P {O}_{4} \left(a q\right) + 2 {H}_{2} O \left(l\right)$

${C}_{4} {H}_{10} \left(g\right) + \frac{13}{2} {O}_{2} \left(g\right) \rightarrow 4 C {O}_{2} \left(g\right) + 5 {H}_{2} O \left(g\right)$

We have thus stoichiometrically balanced an acid-base, and a combustion reaction. $\text{Garbage in equals garbage out.}$

If you like you can double the combustion reaction to remove the non-stoichiometric coefficient in the combustion reaction:

$2 {C}_{4} {H}_{10} \left(g\right) + 13 {O}_{2} \left(g\right) \rightarrow 8 C {O}_{2} \left(g\right) + 10 {H}_{2} O \left(g\right)$

I prefer the former reaction inasmuch as it is easier to specify equivalent masses of hydrocarbon and dioxygen.