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8 July, 16:28

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

2C2H5OH-CH2=C4H6+2H2O+H2

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  1. 8 July, 16:50
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    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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