Question #62c6b

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Answer 1 · 2018-01-15T14:35:03

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

I'm assuming you are talking about complete combustion ("burns completely") of ethane, in that case it will be

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

To balance it, let's balance the oxygens and hydrogens on the right side first.

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

Then, balance the oxygen on the left side.

$C_{2} H_{6} + 3.5 O_{2} \to 2 C O_{2} + 3 H_{2} O$ $C_2H_6+3.5O_2->2CO_2+3H_2O$

There is a fraction with a half, so we can multiply everything by $2$ $2$ .

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

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