Question

6) Ethylene, 15.00 g, is reacted in water, 100.0 mL, and a phosphoric acid catalyst at 250.0°C to give 8.500 g of ethanol. Calculate the percent yield for this reaction?

Answer 1

Avrswer ethanol Ethylene + water - 250C CH₂=CH₂ + H₂O Phospheric acid CH3 - CH2-OH . 8.500g 15.09100mL 100 g woter density u 1g/mL so loomL HO is 100 g H₂O moles of CH = CH 15.04 = 0.534685017 noi 28.053qg|mo moles of H2O = 100 g 100 g 18.0 152g/mo, = 5.550868156 mol reactant So CH₂=CH2 is limiting means H₂o is excess reactant 1 mol CH2=CH₂ is producing 1 mol CH3 -CH2-OH So 0.5346850 17 mol CH2=CH₂ will produce 0.53 4680017 mol ethano) mass of ethanol produced = moles of ethanol x molar mass of ethanol = 0.5346850 17 mox 46.069181mg = 24,63245752 g theoretical field of ethanol = 24.68g actual yield of ethanol = 8.500 g

Perent field = acfucel Yield theorefical field Xloo = 8.500g X100 24.63g = 34.517. Percent yield is 7 34-51%.