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19 February, 21:37

Tetraphosphorus hexaoxide is formed by the reaction of phosphorus with oxygen gas. If a mixture of 75.3 g of phosphorus and 38.7 g of oxygen produce 43.3 g of P4O6, what is the percent yield for the reaction

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  1. 19 February, 23:47
    0
    The percent yield of the reaction is 48.9 %

    Explanation:

    Step 1: data given

    Mass of phosphorus = 75.3 grams

    Molar mass of phosphorus = 123.90 g/mol

    Mass of oxygen = 38.7 grams

    Molar mass oxygen = 32.0 g/mol

    Mass of P4O6 produced = 43.3 grams

    Step 2: The balanced equation

    P4 (s) + 3O2 (g) → P4O6 (s)

    Step 3: Calculate moles P4

    Moles P4 = mass P4 / molar mass P4

    Moles P4 = 75.3 grams / 123.90 g/mol

    Moles P4 = 0.608 moles

    Step 4: calculate moles O2

    Moles O2 = 38.7 grams / 32.0 g/mol

    Moles O2 = 1.21 moles

    Step 5: Calculate the limiting reactant

    For 1 mol P4 we need 3 moles O2 to produce 1 mol P4O6

    O2 is the limiting reactant. There will react 1.21 moles. P4 is in excess. There will react 1.21 / 3 = 0.403 moles. There will remain 0.608 - 0.403 = 0.205 moles P4

    Step 6: Calculate moles P4O6

    For 1 mol P4 we need 3 moles O2 to produce 1 mol P4O6

    For 1.21 moles O2 we'll have 1.21 / 3 = 0.403 moles P4O6

    Step 7: Calculate mass P4O6

    Mass P4O6 = moles P4O6 * molar mass P4O6

    Mass P4O6 = 0.403 moles * 219.88 g/mol

    Mass P4O6 = 88.6 grams

    Step 8: Calculate percent yield

    % yield = (actual yield / theoretical yield) * 100%

    % yield = (43.3 grams / 88.6 grams) * 100 %

    % yield = 48.9 %

    The percent yield of the reaction is 48.9 %
  2. 20 February, 00:54
    0
    percentage yield = 48. 82%

    Explanation:

    Firstly, the chemical equation of the reaction is written and balanced.

    P4 (g) + O2 (g) → P4O6

    balanced equation.

    P4 (s) + 3 O2 (g) → P4O6 (s)

    We have to know the limiting reactant of the reaction by converting the reactants to mole

    molar mass of phosphorus = 4 * 31 = 124 g

    moles of phosphorus = mass/molar mass = 75.3/124 = 0.60725806451 mol-rxn

    molar mass of oxygen = 32 g

    moles of oxygen = 38.7 / 32 = 1.209375 / 3 mol = 0.403125 mol-rxn

    Therefore the limiting reactant is oxygen and it will determine the amount of product

    P4 (s) + 3 O2 (g) → P4O6 (s)

    mass of 3 moles of oxygen = 3 * 32 = 96 g

    molar mass of P4O6 = 124 + 96 = 220 g

    if 96 g of oxygen produces 220 g of P4O6

    38.7 g of oxygen will produce? grams of P4O6

    theoretical yield of P4O6 = (38.7 * 220) / 96

    theoretical yield of P4O6 = 8514 / 96

    theoretical yield of P4O6 = 88.6875

    theoretical yield of P4O6 = 88.69

    percentage yield = actual yield/theoretical yield * 100

    percentage yield = 43.3/88.69 * 100

    percentage yield = 4330/88.69

    percentage yield = 48.8217386402

    percentage yield = 48. 82%
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