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The reaction of 5.3 grams of fluorine with
excess chlorine produced 4.6 grams of ClF3. What percent yield of ClF3 was obtained?
Answer in units of %

  • ap -

    Yield = practical/theoretical

    Given:
    Practical mass: Pm{ClF3} = 4.6g
    Reactant mass: m{F2} = 5.3g
    The equation: 3 F2 + Cl2 = 2 ClF3

    Lookup the Molecular Weights:
    w{F2} = 37.99680650 ± 0.00000002 g/mol
    w{ClF3} = 92.4484 ± 0.0001 g/mol

    So theoretically, the mass of chlorine triflouride is:
    Tm{ClF3} = (2/3) m{F2}w{ClF3}/w{F2}

    Therefore:
    Yield = Pm{ClF3}/Tm{ClF3}
    = (3/2) Pm{ClF3} w{F2} /(m{F2} w{ClF3})
    = (3/2)*4.6*37.99680650/(5.3*92.4484)
    = 53.5%

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