Question #a32cf

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Question #a32cf

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Answer 1 · 2016-03-11T06:58:58

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Explanation:

$2 C_{8} H_{18} + 25 O_{2} = 16 C O_{2} + 18 H_{2} O$ $2C_8H_18+25O_2=16CO_2+18H_2O$

you can remember following also
$C_{x} H_{y} + (x + \frac{y}{4}) O_{2} = x C O_{2} + \frac{y}{2} H_{2} O$ $C_xH_y +(x+y/4)O_2=xCO_2+y/2H_2O$

$\Rightarrow 2 C_{x} H_{y} + (2 x + \frac{y}{2}) O_{2} = 2 x C O_{2} + y H_{2} O$ $=>2C_xH_y +(2x+y/2)O_2=2xCO_2+yH_2O$

Oxidation Number Method
$C_{8} H_{18} + O_{2} \to C O_{2} + H_{2} O$ $C_8 H_18+O_2->CO_2+H_2O$
change in oxidation no of C $\Rightarrow - \frac{9}{4} \to + 4 = \frac{25}{4} u n i t$ $=>-9/4->+4=25/4unit$ oxidation
change in oxidation no of O $\Rightarrow 0 \to - 2 = 2 u n i t$ $=>0->-2=2unit$ Reduction
ratio of no of C and O = $2 : \frac{25}{4} = 8 : 25$ $2:25/4=8:25$
ratio of no of $C_{8} H 18$ $C_8H18$ and O $= 1 : 25$ $=1:25$
ratio of no of $C_{8} H 18$ $C_8H18$ and O $= 2 : 50$ $=2:50$
ratio of no of $C_{8} H 18$ $C_8H18$ and $O_{2} = 2 : 25$ $O_2 =2:25$

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Question #a32cf

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