Balance the following equation by short-cut method, and write the coefficient for
$O_{2}$

$C_{5} H_{11} N H_{2} + O_{2} \to C O_{2} + H_{2} O + N O_{2}$
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Solution

Let the equation have coefficients as follows:
$A C_{5} H_{11} N H_{2} + B O_{2} \to C C O_{2} + D H_{2} O + E N O_{2}$
Balancing the number of atoms of each element on both sides, we get
5A = C
11A + 2A = 2D
13A = 2D
A = E
2B = 2C + 2E + D
$C_{5} H_{11} N H_{2}$ has the largest number of atoms.
Hence let A = E = 1
Then, C = 5
2D = 13 $\Rightarrow$ D = $\frac{13}{2}$
$2 B = 10 + 2 + \frac{13}{2}$
4B = 20 + 4 + 13 = 37
$\Rightarrow B = \frac{37}{4}$
Therefore, $C_{5} H_{11} N H_{2} + (\frac{37}{4}) O_{2} \to 5 C O_{2} + (\frac{13}{2}) H_{2} O + N O_{2}$
Multiplying by 4, we get

$4 C_{5} H_{11} N H_{2} + 37 O_{2} \to 20 C O_{2} + 26 H_{2} O + 4 N O_{2}$

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