Calculate the mass of ethyl alcohol required to prepare 540 grams of C4H6, if the reaction follows the scheme:

2C2H5OH-CH2=C4H6+2H2O+H2

2C₂H₅OH = C₄H₆ + 2H₂O + H₂

2 mole 1 mole

molecular weight of ethyl alcohol

mol weight of C₂H₅OH = 46 gm

mol weight of C₄H₆ 54 gm

540 gm of C₄H₆ = 10 mole

10 mole of C₄H₆ will require 20 mol of ethyl alcohol.

20 mole of ethyl alcohol = 20 x 46

= 920 gm

ethyl alcohol required = 920 gm.