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18 September, 12:39

Bromine (63 g) and fluorine (60 g) are mixed to give bromine trifluoride. a) Write a balanced chemical reaction. b) What is the limiting reactant? c) Determine how many grams of the product will form if the percent yield of the reaction is 80%. d) Determine how many grams of the excess reactant is left unreacted.

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  1. 18 September, 12:57
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    a) Br2 + 3F2 → 2BrF3

    b) Br2 is the limiting reactant.

    c) There will be formed 86.3 grams of BrF3

    d) There will remain 0.326 moles of F2 = 12.97 grams

    Explanation:

    Step 1: The balanced equation

    Br2 + 3F2 → 2BrF3

    Step 2: Given data

    Mass of Bromine = 63grams

    Mass of fluorine = 60 grams

    percent yield = 80%

    Molar mass of bromine = 79.9 g/mol =

    Molar mass of fluorine = 19 g/mol

    Molar mass of bromine trifluoride = 136.9 g/mol

    Step 3: Calculating moles

    Moles Br2 = 63 grams / (2*79.9)

    Moles Br2 = 0.394 moles

    Moles F2 = 60 grams / (2*19.9)

    Moles F2 = 1.508

    For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3

    Br2 is the limiting reactant. It will completely be consumed (0394 moles).

    There will react 3*0.394 = 1.182 moles of F2

    There will remain 1.508 - 1.182 = 0.326 moles of F2 = 12.97 grams

    Step 5: Calculate moles of BrF3

    For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3

    So there is 2*0.394 moles = 0.788 moles of BrF3 moles produced

    Step 6: Calculate mass of BrF3

    mass = Moles * Molar mass

    mass of BrF3 = 0.788 moles * 136.9 g/mol = 107.88 grams = Theoretical yield

    Step 7: Calculate actual yield

    % yield = 0.80 = actual yield / theoretical yield

    actual yield = 0.80 * 107.88 grams = 86.3 grams

    actual yield = 86.3 grams
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