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12mll OF A GASEOUS HYDROCARBON WAS EXPLODED WITH 50ml OF OXYGEN .THE VOLUME AFTER EXPLOSION WAS 32 ml. ON TREATMENT WITH KOH THE VOLUME REDUCED TO 8ml. THEN FORMULA OF HYDROCARBON IS - C
_{2}H_{2} - C
_{2}H_{6} - C
_{3}H_{8}

_{2}H_{2}_{2}H_{6}_{3}H_{8}_{x}H

_{y}+ O

_{2}$\to $CO

_{2}+ H

_{2}O

Volume of (CO

_{2}

_{ }+ unreacted O

_{2}) = 32 ml.

As CO

_{2}can be absorbed by KOH . When this 32 ml. mixture is treated with KOH, it is reduced to 8 ml.

Therefore amount of CO

_{2}formed = 32 ml. - 8ml.= 24 ml.

Amount of unreacted O

_{2}= 8 ml.

Volume of O

_{2}reacted = 50 ml - 8 ml = 42 ml.

Volume of hydrocarbon = 12 ml.

Apply Principle of Atom Conservation (POAC) on C we get:

X

_{}moles of C

_{x}

_{ }H

_{y}= 1

_{}no. of moles of CO

_{2}

X $\times $12 ml = 1 x 24 ml

**X = 24/12 = 2**Apply POAC on H atoms

Y

_{$\times $ }no. of moles of C_{x}H_{y}= 2_{$\times $}no. of mole of H_{2}O -----------------( 1 )Apply POAC on O atoms

2

_{}no. of moles of O_{2}= 1^{$\times $}no. of mole of H_{2}O + 2^{$\times $}no. of mole of CO_{2}From eq. (1)

2

_{}no. of moles of O_{2}= y/2^{$\times $}no. of mole of C_{x}H_{y}+ 2^{$\times $}no. of mole of CO_{2}2$\times $42 = Y/2$\times $12 + 2 *24

84 = 6Y + 48

84-48 = 6Y

36 = 6Y

So the formula of hydrocarbon =

84-48 = 6Y

36 = 6Y

**Y = 6**So the formula of hydrocarbon =

**C**_{2}H_{6}
**
**