C\(_x\)H\(_y\) + 9O\(_2\) \(\rightarrow\) 6CO\(_2\) + 6H\(_2\)O The hydrocarbon, C\(_x\)H\(_y\), in the reaction above is...

Explanation

To determine the type of hydrocarbon (C_xH_y) involved in the reaction, we can first balance the equation and find the general formula of the hydrocarbon. The balanced equation is:

\(C_xH_y + 9O_2 \rightarrow 6CO_2 + 6H_2O\)

Now, we can equate the number of carbon and hydrogen atoms on both sides of the equation:

Carbon atoms: 6 = x

Hydrogen atoms: 12 = y

So, the hydrocarbon has the formula C₆H₁₂. Now, let's consider the types of hydrocarbons given in the options:

Option A: Alkane - Alkanes have the general formula C_nH_2n+2. In this case, C₆H₁₂ does not fit the formula for an alkane.

Option B: Benzene - Benzene has the formula C₆H₆, which is not the same as the given hydrocarbon.

Option C: Alkene (Correct) - Alkenes have the general formula C_nH_2n. C₆H₁₂ fits this formula, so the hydrocarbon is most likely an alkene.

Option D: Alkyne - Alkynes have the general formula C_nH_2n-2. C₆H₁₂ does not fit the formula for an alkyne.

In conclusion, the hydrocarbon in the reaction is most likely an alkene, which is consistent with the correct option (C).