# 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​2H2 C2H6 C3H8

CxHy + O2 $\to$CO2 + H2O
​Volume of (CO2 + unreacted O2) = 32 ml.
As CO2  can be absorbed by KOH .  When this 32 ml. mixture is treated with KOH, it is reduced to 8 ml.
Therefore amount of CO2 formed = 32 ml. - 8ml.= 24 ml.
Amount of unreacted O2 = 8 ml.
Volume of O2 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 Cx Hy = 1 no. of moles of CO2
$×$12 ml = 1 x 24 ml
X = 24/12 = 2
Apply POAC on H atoms
$×$ no. of moles of Cx Hy = 2$×$ no. of mole of H2O -----------------( 1 )
Apply POAC on O atoms no. of moles of O2 = 1$×$ no. of mole of H2O + 2 $×$ no. of mole of CO2
From eq. (1) no. of moles of O2 = y/2$×$ no. of mole of Cx Hy + 2 $×$ no. of mole of CO2
2$×$42 = Y/2$×$12 + 2 *24
84 = 6Y + 48
84-48 = 6Y
​36 = 6Y
Y = 6
So the formula of hydrocarbon = C2H6

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