92 g of ethanol was passed over heated alumina to obtain 40 liters of ethylene. What is the ethylene yield in% of the theoretical possible?
Let’s write the reaction equation:
C2H5OH (t, Al2O3) = C2H4 + H2O
Let’s find the amount of the substance C2H5OH:
v (C2H5OH) = m (C2H5OH) / M (C2H5OH) = 92/46 = 2 (mol).
According to the reaction equation, 1 mol of C2H4 is formed from 1 mol of C2H5OH, therefore:
v (C2H4) = v (C2H5OH) = 2 (mol).
Thus, at 100% reaction yield, the volume of ethylene obtained, measured under normal conditions (n.o.):
Vtheor. (C2H4) = v (C2H4) * Vm = 2 * 22.4 = 44.8 (l).
Let’s find the ethylene yield in this reaction:
Ф (C2H4) = (V (C2H4) / Vtheor. (C2H4)) * 100 = (40 / 44.8) * 100 = 89.286 = 89 (%).
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