A gaseous mixture of methane ( CH 4 ) and ethane ( C 2 H 6 ) has a total volume of 10.0 L at standard temperature and pressure (STP). When this mixture undergoes complete combustion, it produces

Question

A gaseous mixture of methane ( CH 4 ​ ) and ethane ( C 2 ​ H 6 ​ ) has a total volume of 10.0 L at standard temperature and pressure (STP). When this mixture undergoes complete combustion, it produces 16.0 L of carbon dioxide ( CO 2 ​ ) at STP. What is the volume percentage of methane in the original mixture?

TestHub · Accepted Answer

The combustion reactions are:

1. CH₄(g) + 2O₂(g) $\to$ CO₂(g) + 2H₂O(l)

2. 2C₂H₆(g) + 7O₂(g) $\to$ 4CO₂(g) + 6H₂O(l)

Let $V_{CH_{4}}$ and $V_{C_{2} H_{6}}$ be the volumes of methane and ethane.

Total volume: $V_{CH_{4}} + V_{C_{2} H_{6}} = 10.0 L$ (Eq. 1)

From reaction 1, 1 L CH₄ produces 1 L CO₂.

From reaction 2, 1 L C₂H₆ produces 2 L CO₂.

Total CO₂ produced: $V_{CH_{4}} + 2 V_{C_{2} H_{6}} = 16.0 L$ (Eq. 2)

Subtracting Eq. 1 from Eq. 2 gives $V_{C_{2} H_{6}} = 6.0 L$ .

Substituting into Eq. 1: $V_{CH_{4}} + 6.0 = 10.0 \Rightarrow V_{CH_{4}} = 4.0 L$ .

Volume percentage of methane = $(4.0 L /10.0 L) \times 100% = 40%$ .

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