Answer to Question #155347 in Inorganic Chemistry for Okonji victor

Question #155347

5mol of a mixture of CH4 and C2H6 requires exactly 358.4L of oxygen for complete combustion, determine the percentage of methane in the mixture (GMV =22.4,dm³ at s.t.p


1
Expert's answer
2021-01-14T04:25:23-0500

Mole = 358./4dm3/ 22.4dm3

= 16gmol-1

Total mole = 21gmol-1

Pecentage of methane© = 5gmol-1 /21gmol-1

=0.2%


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