A certain hydrocarbon on complete combustion at s.t.p produced 89.6dm(^3) of CO(_2) and 54g of water. The hydrocarbon should be
-
A.
C(_6)H(_6) -
B.
C(_4)H(_{10}) -
C.
C(_5)H(_{10}) -
D.
C(_4)H(_6)
Correct Answer: Option D
Explanation
C(_x)H(_y) + (x + y/4)O(_2) (to) x CO(_2) + y/2 H(_2)O
89.6 dm(^3) (to) x moles CO(_2)
22.4 dm(^3) (to) 1 mole CO(_2)
x = (frac{89.6}{22.4})
= 4 moles CO(_2)
54g of H(_2)O (to) k moles H(_2)O
18g H(_2)O (to) 1 mole H(_2)O
k = (frac{54}{18})
= 3 moles H(_2)O
From the equation above,
x = 4
k = y/2 = 3
y/2 = 3 (implies) y = 6
(therefore) The hydrocarbon C(_x)H(_y) = C(_4)H(_6)
There is an explanation video available below.