What are the coefficients of all in lowest term after balancing the reaction?
$. . . C_{4} H_{10} + . . . . O_{2} \to . . . . C O_{2} + . . . . H_{2} O$

1, 6, 4, 2

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2, 13, 8, 10

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1, 6, 1, 5

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3, 10, 16, 20

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4, 26, 16, 20

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Solution

The correct option is A $2, 13, 8, 10$
The unbalanced chemical equation is $C_{4} H_{10} + O_{2} \to C O_{2} + H_{2} O$
Balance C atoms
$C_{4} H_{10} + O_{2} \to 4 C O_{2} + H_{2} O$
Balance H atoms
$C_{4} H_{10} + O_{2} \to 4 C O_{2} + 5 H_{2} O$
Balance O atoms
$C_{4} H_{10} + \frac{13}{2} O_{2} \to 4 C O_{2} + 5 H_{2} O$
This is the balanced chemical equation.
Multiply above equation with 2 so that we have whole numbers as coefficients.
$2 C_{4} H_{10} + 13 O_{2} \to 8 C O_{2} + 10 H_{2} O$
The coefficients of all in lowest term after balancing the reaction are $2, 13, 8, 10$

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