Question

Question #a32cf

Answer 1

See below

Explanation:

`2C_8H_18+25O_2=16CO_2+18H_2O`

you can remember following also
`C_xH_y +(x+y/4)O_2=xCO_2+y/2H_2O`

`=>2C_xH_y +(2x+y/2)O_2=2xCO_2+yH_2O`

Oxidation Number Method
`C_8 H_18+O_2->CO_2+H_2O`
change in oxidation no of C `=>-9/4->+4=25/4unit` oxidation
change in oxidation no of O `=>0->-2=2unit` Reduction
ratio of no of C and O =`2:25/4=8:25`
ratio of no of `C_8H18` and O`=1:25`
ratio of no of `C_8H18` and O`=2:50`
ratio of no of `C_8H18` and`O_2 =2:25`