When the equation: $C_{2} H_{6} + O_{2} \to C O_{2} + H_{2} O$ is completely balanced using the lowest whole number coefficients, the sum of the coefficients will be :

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Solution

The correct option is C 19
The unbalanced equation is
$C_{2} H_{6} + O_{2} \to C O_{2} + H_{2} O$
Balance C atoms
$C_{2} H_{6} + O_{2} \to 2 C O_{2} + H_{2} O$
Balance H atoms
$C_{2} H_{6} + O_{2} \to 2 C O_{2} + 3 H_{2} O$
Balance O atoms
$C_{2} H_{6} + \frac{7}{2} O_{2} \to 2 C O_{2} + 3 H_{2} O$
Multiply the above equation with 2 to get whole number coefficients.
$2 C_{2} H_{6} + 7 O_{2} \to 4 C O_{2} + 6 H_{2} O$
This is the balanced equation.
The sum of the coefficients will be $2 + 7 + 4 + 6 = 19$ .

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