oao822 wrote:I have a flow of 50 t/h of C2H6 that contains 1mol% of C2H4
How would i convert this 1 mol% into T/H.
Thanks
First note that you have 50 t/h of
mixture.
The molar mass of C2H6 is 30.068 g and C2H4 is 28.052 g
The 1% mole fraction mixture means that 100 mol of mixture is
99 moles * 30.068 g/mol C2H6 + 1 mol * 28.052 g/mol C2H4 = 3004.784 g mixture.
By weight, the mixture is 28.062 g C2H4/3004.784 g mixture or 0.9338%. Multiply by 50 t/h, and you have 0.4668 t/h C2H4 (and 49.533 t/h of C2H6)
