Question #dc820

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Question #dc820

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Answer 1 · 2015-05-09T21:42:45

A combustion reaction will only produce carbon dioxide, $C O_{2}$ $CO_2$ , and water, $H_{2} O$ $H_2O$ . An unbalanced chemical equation for the combustion of a general hydrocarbon, $C_{x} H_{y}$ $C_xH_y$ , wil be

$C_{x} H_{y} + O_{2} \to C O_{2} + H_{2} O$ $C_xH_y + O_2 -> CO_2 + H_2O$

Regardless of what values you have for $x$ $x$ and $y$ $y$ , the balanced chemical equation will always have these stoichiometric coefficients

$C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to x C O_{2} + \frac{y}{2} H_{2} O$ $C_xH_y + (x+y/4)O_2 -> xCO_2 + y/2H_2O$

Take, for example, the combustion of methane, $C H_{4}$ $CH_4$ . For methane, you have $x = 1$ $x=1$ and $y = 4$ $y=4$ . PLug these values into the equation above and you'll get

$C_{1} H_{4} + (1 + \frac{4}{4}) O_{2} \to C O_{2} + \frac{4}{2} H_{2} O$ $C_1H_4 + (1 + 4/4)O_2 ->CO_2 + 4/2H_2O$

which ends up being

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$ $CH_4 + 2O_2 -> CO_2 + 2H_2O$ $\to$ $->$ balanced

Another example could be the combustion of ethane, $C_{2} H_{6}$ $C_2H_6$ , for which $x = 2$ $x=2$ and $y = 6$ $y=6$ .

$C_{2} H_{6} + (2 + \frac{6}{4}) O_{2} \to 2 C O_{2} + \frac{6}{2} H_{2} O$ $C_2H_6 + (2 + 6/4)O_2 -> 2CO_2 + 6/2H_2O$

which ends up being

$C_{2} H_{6} + \frac{7}{2} O_{2} \to 2 C O_{2} + 3 H_{2} O$ $C_2H_6 + 7/2O_2 -> 2CO_2 + 3H_2O$

If you're not allowed to use fractional coefficients, multiply all the coefficients by 2 to get

$2 C_{2} H_{6} + 7 O_{2} \to 4 C O_{2} + 6 H_{2} O$ $2C_2H_6 + 7O_2 -> 4CO_2 + 6H_2O$ $\to$ $->$ balanced

That's how you balance the combustion reaction of any hydrocarbon.

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Question #dc820

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