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\(_6\)
jamb 2019chemistry

The correct answer is D. C\(_4\)H\(_6\)

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\(_6\)

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