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A natural gas may be assumed to be a mixture of CH_(4) and C_(2)H_(6) only. On complete combination of 10 L of the gas at STP, the heat evolved was 474.6 kJ. Assuming DeltaH_(c)(CH_(4)(g)) = -894 kJ/mol and DeltaH_(c)(C_(2)H_(6)(g)) = -1560 kJ//mol. Calculatethe % of volume of each gas in the mixture. |
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Answer» Solution :`K_(s) to K(G), DeltaH_(5) = 89.8 KJ mol^(-1)` We have to find `DELTAH` of the followingequation. |
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