The complete combustion of 1.5 x 10^-3 moles of a gaseous hydrocarbon required 84 cm³ of oxygen (measured at S.T.P) and produced 27 mg of water - Leaving Cert Chemistry - Question b - 2001

To establish that the hydrocarbon is ethyne (C₂H₂), we can analyze the combustion reaction:

For the combustion of C₂H₂:

$C_2H_2 + 2 O_2 \rightarrow 2 CO_2 + H_2O$

Using the stoichiometry:

• 2 moles of oxygen are needed to combust 1 mole of C₂H₂. Given: 0.00375 moles of O₂ → Using the ratio: $\frac{1 ext{ mol } C_2H_2}{2 ext{ mol } O_2}$ We calculate moles of C₂H₂: $\text{Moles of } C_2H_2 = \frac{0.00375}{2} = 0.001875 ext{ moles}$

Next, calculating moles of produced CO₂ and H₂O: Given total moles of products:

• Moles of CO₂ produced = 0.001875 imes 2 = 0.00375 moles.
• Moles of water produced = 0.001875 moles.

Carbon to hydrogen ratio confirms the empirical formula is C₂H₂, demonstrating the hydrocarbon is indeed ethyne.

The product of the hydration of ethyne (C₂H₂) is ethanol (C₂H₅OH) or acetaldehyde.

Reagents and Conditions:

1. Reagents: The hydration reaction typically requires water (H₂O) and a catalyst such as dilute sulfuric acid (H₂SO₄) or mercury(II) sulfate (HgSO₄).
2. Conditions: The reaction conditions involve heating at temperatures generally between 40°C and 100°C, with sufficient pressure to favor the addition reaction.