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11 June, 04:30

Given 6.0 mol of N2 are mixed with 12.0 mol of H2. Equation: N2 + 3H2--> 2NH3

Which chemical is in excess? What is the excess in moles?

Theoretically, how many moles of NH3 will be produced?

If the percentage yield of NH3 is 80%, how many moles of NH3 are actually produced?

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  1. 11 June, 04:59
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    1) Excess reagent

    1 mol N2 / 3 mol H2

    6.0 mol N2 * 3 mol H2 / 1 mol N2 = 18 mol H2

    18mol H2 > 12 mol H2 = > H2 is limiting (you need 18 mol H2 to use all the 6 mol N2), then N2 is in excees.

    12.0 mol H2 * 1mol N2 / 3 mol H2 = 4 mol N2 is the quantity that will react, then the excess is 6 mol N2 - 4 mol N2 = 2 mol N2

    2) NH3 produced

    12 mol H2 * [2 mol NH3 / 3 mol H2] = 8 mol NH3

    Aslso, 4 mol N2 * [2molNH3 / 1 molN2] = 8 mol NH3, the same result.

    3) Yield

    80% * 8 mol NH3 = 6.4 mol NH3
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