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