Question

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

Answer 1

Mole = 358./4dm³/ 22.4dm³

= 16gmol^-1

Total mole = 21gmol^-1

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

=0.2%