The complete combustion of 1.5 x 10² 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 demonstrate that the hydrocarbon is ethyne (C₂H₂), we first refer to the general combustion reaction of hydrocarbons:

$C_nH_m + O_2 \rightarrow CO_2 + H_2O$

Assuming the combustion product aligns with this reaction, we have:

For every 1 mole of C₂H₂, we require 2 moles of O₂ and produce 2 moles of CO₂ and 1 mole of H₂O. Matching the known values:

• Moles of oxygen used calculated previously: 0.00375 Moles
• Moles of water produced: 0.0015 Moles

With the stoichiometry of ethyne:

• Oxygen: 2 moles needed for 1 mole of ethyne → 0.00375 moles implies approximately 0.001875 moles of C₂H₂.
• Water: For 1 mole of ethyne, we produce 1 moles of water; hence 0.0015 moles of water corresponds to approximately 0.0015 moles of ethyne.

Thus, the hydrocarbon meets the combustion requirement for ethyne.

The hydration of ethylene (C₂H₄) results in the formation of ethanol (C₂H₅OH). This reaction typically requires specific conditions and reagents:

1. Reagent: Phosphoric acid (H₃PO₄) or sulfuric acid (H₂SO₄) is often used as a catalyst in the reaction.
2. Conditions: The process is usually carried out at elevated temperatures (around 300°C) and under high pressure.

The overall reaction can be represented as:

$C_2H_4 + H_2O \xrightarrow{H_3PO_4} C_2H_5OH$

In summary, ethanol is the product of ethylene hydration, facilitated by a strong acid catalyst under specific temperature and pressure conditions.