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

Chemistry

1 answer:

kkurt [141]3 years ago

3 0

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 .

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