1.

44g of a sample on complete combusion given 88 gm CO_2 and 36 gm of H_2O. The molecular formula of the compound may be :-

Answer»

`C_2H_6`
`C_2H_6O`
`C_2H_4O`
`C_3H_6O`

SOLUTION :`C_xH_y+((2x+y//2)/2)O_2 to xCO_2 + y//2H_2O`
`{:(44G,"88G"/44,36/18),(,2,2):}`
x=2
y/2=4
x=2
y=4


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