A natural gas may be assumed to be a mixture of methane and ethane only. On complete combustion of `10L` of gas at `STP` the heat evolved was `474.6 k

Question

A natural gas may be assumed to be a mixture of methane and ethane only. On complete combustion of `10L` of gas at `STP` the heat evolved was `474.6 k

A natural gas may be assumed to be a mixture of methane and ethane only. On complete combustion of `10L` of gas at `STP` the heat evolved was `474.6 kJ`. Assuming `Delta_(comb) H^(Theta) CH_(4)(g) =- 894 kJ mol^(-1)` and `Delta_(comb) H^(Theta) C_(2)H_(6)(g) =- 1500 kJ mol^(-1)` . So, find composition of the mixture by volume.

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

x litre `toCH_(4)," mole of "CH_(4)=x//22.4`
`(10-x)litre to C_(2)H_(6)," mole of "C_(2)H_(6)=(10-x)//22.4`
Heat evolved `=(x)/(22.4)xx894+((10-x))/(22.4)xx1500`
`474.6=(x)/(22.4)xx894+((10-x))/(22.4)xx1500`
`x=0.745,%CH_(4)=74.5%`

4 views Answered 2023-02-05 15:29