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\(_6\)
Verified Correct Answer: D.
Step-by-Step Mathematical Solution
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\(_6\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #55521)