Question

ment-take For the following reaction, 5.59 grams of hydrogen gas are allowed to react with 10.7 grams of ethylene (C2H) hydrogen (g) + ethylene (C2H) (g) ethane (C2H (g) What is the maximum amount of ethane (C,H) that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

5.59y = 2,773 mol H2C=CH2(0)+ H21) Hoc-CH3 (9) 5.59 g hydrogen gas = 2016 glinor = 21 10.74 eltelene u = 105 1054 ) = 0.381 mol of moles ethiefene is less. so elebene . Here no of I gik Ha Now, is the limiting reagent. I mote ethetene react react with with - to powdhe I more ethane. - sv. from 0. ggr mot ethelere oto & we will get 0.381 mol ethane. o oo Moreimum amount of ether can be formed !!! . . = 0.381 mol = 0.381 mblx 30.079 por = 11:469 Formule of limiting reagent = ((242) Excess reyent = (2-713 mol - 0.381 mol ) 2. 392 mol = 12.392 X 2.016)g - 41829