How do you balance these equations?

Question

$H_{3} P O_{4} (a q) + M {g (O H)}_{2} (a q) \to M g_{3} {(P O_{4})}_{2} (a q) + H_{2} O (l)$ $H_3PO_4 (aq) + Mg(OH)_2(aq) -> Mg_3(PO_4)_2(aq) + H_2O(l)$
$C_{4} H_{10} (g) + O_{2} (g) \to C O_{2} (g) + H_{2} O (g)$ $C_4H_10(g) + O_2 (g) -> CO_2 (g) + H_2O(g)$

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Answer 1 · 2016-12-03T06:49:44

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

$H_{3} P O_{4} (a q) + M {g (O H)}_{2} (a q) \to M g H P O_{4} (a q) + 2 H_{2} O (l)$ $H_3PO_4(aq) + Mg(OH)_2(aq) rarr MgHPO_4(aq) + 2H_2O(l)$

$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (g)$ $C_4H_10(g) + 13/2O_2(g) rarr 4CO_2(g) + 5H_2O(g)$

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

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

$2 C_{4} H_{10} (g) + 13 O_{2} (g) \to 8 C O_{2} (g) + 10 H_{2} O (g)$ $2C_4H_10(g) + 13O_2(g) rarr 8CO_2(g) + 10H_2O(g)$

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