Answer:
<u>Percentage Yield = 93.8%</u>
Explanation:
<u>Data:</u>
R = 0.08206 L.atm/mol/K,
Molar mass of C2H6 = 30.08 g/mol
T = 300° c +273K = 573.15 K
C2H4 Conditions,
P = 25 atm
V = 1000 L/min
H2 Condition,
P = 25 atm
V = 1500 L/min
<u>Solution:</u>
First we will calculate the moles of ethane produced in one minute,
nC2H6 = (15/30.08) ( 1000/1)
<u>nC2H6 = 498.67 mol/min</u>
Now we will calculate the moles of ethane flowing in one minute using the ideal gas equation which is, Pv = nRT
n = pV/R
nC2H4 = ((25* 1000)/(0.08206*573.15))
<u>nC2H4 = 531.54 mol/min</u>
Similarly Calculating the moles of hydrogenflowing in one minute,
nH2 = ((25*1500)/(0.08206)(573.15))
<u>nH2 = 797.32 mol/mim</u>
Both hydrogen and ethene are flowing at 25 atm and 573.15 K but the flow rate of hydrogen is more than that of ethane which results in less amount of available ethene for the reaction. We know that 1 mole of ethene reacts with 1 mole of hydrogen to give 1 mole of ethane
<u>hence ethene is acting as the limiting reagent here </u>
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Now calculating the %Yield = (498.67/531.54) * 100
<u> %Yield = 93.82%</u>