$C_{5} H_{11} {NH}_{2} + O_{2} \to {CO}_{2} + H_{2} O + {NO}_{2}$

The coefficient of $O_{2}$ after balancing the above equation is _____.

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Solution

The correct option is B 37
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
$5 A = C$
$11 A + 2 A = 2 D$
$13 A = 2 D$
$A = E$
$2 B = 2 C + 2 E + D$

$C_{5} H_{11} N H_{2}$ has the largest number of atoms. Hence, let $A = E = 1$
Then, $C = 5$
$2 D = 13$ $\Rightarrow D = \frac{13}{2}$
$2 B = 10 + 2 + \frac{13}{2}$
$4 B = 20 + 4 + 13 = 37$ $\Rightarrow B = \frac{37}{4}$

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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