Question

A gaseous mixture of propane, acetylene and `CO_(2)` is burnt in excess of air. Total `4800` KJ heat is evolved . The volume of `CO_(2)` formed during combustion is `224` liters at STP.
The total evolved heat is used to perform two separate process:
`(i)` Vapourising `87.5%` of water (liquid) obtained in the process of burning the original mixture.
`(ii)` Forming `3808` litres ethylene measured at STP from its elements.
`DeltaH_(H-H) = 435 KJ//mol " " DeltaH_(C-H) = 416 KJ//mol " " DeltaH_(C-C) = 347 KJ//mol`
` DeltaH_(C-C) = 615 KJ//mol, " " DeltaH_(C-C) = 812 KJ//mol" " DeltaH_("sublimation") "of " (C,s) = 718 KJ//mol`
`DeltaH_(f)^(@)(C_(2)g) = -394 KJ//mol " "DeltaH_(f)^(@)(H_(2)O, l) = -286 KJ//mol. " " DeltaH_(f)^(@)(H_(2)O,g) = -247 KJ//mol.`
Total moles of hydrocarbon gases taken in the initial mixture
A. `3`
B. `4`
C. `2`
D. `5`

Answer 1

Correct Answer - A
Let moles of `C_(3)H_(8) = X , " " "moles of" C_(2)H_(2) = y`
& moles of `CO_(2) =Z.`
Calculation of `DeltaH_(f)^(@) "of" C_(3)H_(8) (g)`
`" "3C(s) + 4 H_(2) (g) rarr C_(3)H_(8)(g)`
`DeltaH_(f)^(@) "of" C_(3)H_(8)(g) =[3(718) + 4(435)] - [2(347) + 8(416)]`
`= 3894- 4022 =-128KJ//mol`
Calculation of `DeltaH_(f)^(@) "of" C_(2)H_(2) (g)`
`" "2C(s) + H_(2) (g) rarr C_(2)H_(2)(g)`
` DeltaH_(f)^(@) C_(2)H_(2)(g)=[2(718) + (435)] - [(812) + 2(416)]`
`=(1436 + 435) - [1644]`
`=227 KJ//mol`.
Calculation of `DeltaH_(f)^(@) "of" C_(2)H_(4) (g)`
`2C(s) + 2H_(2) (g) rarr C_(2)H_(4)(g)`
` DeltaH_(f)^(@) "of" C_(2)H_(4)(g) = [2(718) + 2(435)] - [615 + 4(416)]`
`= 2306 - 2279 = 27 KJ//mol`.
Calculation of `DeltaH_("Comb")^(@) "of" C_(3)H_(8)(g)`
`" "C_(3)H_(8)(g) + O_(2)(g) rarr 3CO_(2)(g) + 4H_(2)O(l)`
`" " DeltaH_("Comb")^(@) "of" C_(3)H_(8) = [3 DeltaH_(f^(@))(CO_(2)) + 4Delta_(F^(@)) (H_(2)O,l)] - DeltaH_(F^(@)) (C_(3)H_(8),g)`
`" " = [3(-394)+ 4(-286)] - (-128) =-2198 KJ//mol`.
Calculation of `DeltaH_("Comb")^(@) "of" 2C_(2)H_(2)(g)`
`C_(2)H_(2)(g) + 2.50_(2)(g) rarr 2CO_(2)(g) + H_(2)O(l)`
`DeltaH_("Comb")^(@) " of " C_(2)H_(2) = [2DeltaH_(f^(@)) (CO_(2)) + DeltaH_(F^(@))(H_(2)O, l)] - Delta_(F^(@))(C_(2)H_(2))`
`=[2(-394)+ (-286)] - 227`
`=-1301 KJ//mol.`
Now total heat released during combustion
`" " 2198 x + 1301 y = 4800" "(i)`
`" "` Combustion `=3x + 2y + z =10 " "(ii)`
Total moles of `H_(2)O(l) "formed" = 4x + y.`
moles of `C_(2)H_(4)(g)` to be prepared `=(3808)/(22.4) = 170.`
Total heat absorbed during evaporation of water formation of `170` moles `C_(2)H_(4)= 4800.`
`" " [(4x + y ) xx 0.875 xx 40] + (170 xx 27) = 4800`
`" " 4x + y =6" (iii)`
on solvent `(i), (ii)` and `(iii)`
we get `x=1, y=2 "and" z=3`