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11 October, 09:15

14.0g N2 react with 3.15g H2 to give an actual yield of 14.5 NH3

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  1. 11 October, 13:07
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    You can determine the limiting reactant and the percentaje yield as follow.

    1) State the equation that represents the chemical reaction

    N2 (g) + 3H2 (g) → 2NH3 (g)

    molar ratios: 1 mol N2 : 3 mol H2 : 2 mol NH3

    2) Convert the two masses to moles by dividing each by its molar mass

    14.0 g N2 / 28 g/mol = 0.5 mol

    3.15 g H2 / 2g/mol = 1.575 mol

    Proportion = 0.5 mol N2 / 1.575 mol H2 = 0.317 mol N2 / mol H2

    3) Given that the theoretical ratio 1 / 3 > actual proportion 0.317, the H2 is the limiting reactant (it will be consumed and yet some moles of N2 will remain without reacting).

    4) Percentage yield = actual yield * 100 / theoretical yield

    Theoretical yield = [ 2 mole NH3 * 1 mole N2 ] * 0.5 mol N2 = 1 mole NH3

    actual yield = 14.5 g NH3 / molar mass NH3 = 14.5 g / 17 g/mol = 0.853 mol

    Percentage yield = actual yield * 100 / theoretical yield = 0.853 mol * 100 / 1 mol = 85.3%

    Answer: 85.3%
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