Question

A natural gas may be assumed to be a mixture fo methane and ethane only. On complete combustion of 10L of gas at STP the heat evolved was 474.6kJ. Assuming ΔₒₘᵦHθCH₄(g)=−894kJmol⁻¹ and composition of the mixture by volume.

Answer 1

The heat evolved from the complete combustion of 10L of natural gas at STP is 474.6 kJ. Assuming the natural gas is a mixture of methane and ethane only, the composition of methane in the mixture by volume is 5.32%.

Step-by-step explanation:

The heat evolved from the complete combustion of 10L of natural gas at STP is 474.6 kJ. We can assume that the natural gas is a mixture of methane and ethane only. The thermochemical equation for the combustion of methane is CH₄(g) + 2O₂ (g) → CO₂ (g) + 2H₂O (1) + 890.4 kJ. From this equation, we can see that the combustion of 1 mol of methane releases 890.4 kJ of heat energy.

In this case, we can use stoichiometric calculations to find the amount of methane that was combusted. Since the reaction of 1 mol of methane releases 890.4 kJ, the combustion of 0.532 mol of methane would release 890.4 kJ. This means that the volume of methane combusted is 0.532 L.

To determine the composition of the mixture by volume, we divide the volume of methane (0.532 L) by the total volume of the gas (10 L) and multiply by 100. Therefore, the composition of methane in the mixture by volume is 5.32%.