$C_{4} H_{10} (g) + O_{2} (g) \to C O_{2} (g) + H_{2} O (l)$
Find the coefficient of $C O_{2}$ , when the above equation is balanced using the lowest whole-number terms?

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Solution

The correct option is E 8
The unbalanced chemical equation is $C_{4} H_{10} (g) + O_{2} (g) \to C O_{2} (g) + H_{2} O (l)$
On LHS, we have $4 C$ atoms. So on RHS we add coefficent of 4 to $C O_{2}$ to balance $C$ atoms.
$C_{4} H_{10} (g) + O_{2} (g) \to 4 C O_{2} (g) + H_{2} O (l)$
On LHS, we have $10 H$ atoms. To balance $H$ atoms, we add coefficient 5 to $H_{2} O$
$C_{4} H_{10} (g) + O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l)$
On LHS, we have $2 O$ atoms. On RHS, we have $13 O$ atoms. To balance $O$ atoms, we add $\frac{13}{2}$ to $O_{2}$ on LHS
$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l)$
This is balanced chemical equation. For using the lowest whole-number terms, we multiply this equation with 2.
$2 C_{4} H_{10} (g) + 13 O_{2} (g) \to 8 C O_{2} (g) + 10 H_{2} O (l)$
The coefficient of $C O_{2}$ in the above equation is 8.

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