Question

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

Answer 1

Answer:

Explanation:

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 .