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15 January, 11:10

A chemical engineer calculated that 15.0 mol H2 was needed to react with excess N2 to prepare 10.0 mol NH3. But the percentage yield of the reaction is 60.0%.

a. Write a balanced equation for the reaction

b. How many moles of NH3 were actually made based on the percentage yield?

c. How many grams of NH3 are formed?

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  1. 15 January, 11:31
    0
    A. 3H2 + N2 - > 2NH3

    B. 6 moles

    C. 170g

    Explanation:

    A. The balanced equation for the reaction? This is given below:

    3H2 + N2 - > 2NH3

    B. Determination of the actual yield of NH3. This is illustrated below:

    Theoretical yield = 10 moles

    Percentage yield = 60%

    Actual yield = ?

    Percentage yield = Actual yield/Theoretical yield x100

    60/100 = Actual yield / 10

    0.6 = Actual yield/10

    Cross multiply

    Actual yield = 0.6 x 10

    Actual yield of NH3 = 6 moles

    C. Determine the mass of NH3 produced.

    Number of mole NH3 = 10 moles

    Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol

    Mass of NH3 = ?

    Mass = number of mole x molar Mass

    Mass of NH3 = 10 x 17

    Mass of NH3 = 170g
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