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7 July, 13:51

Identify the limiting reactant in the reaction of bromine and chlorine to form BrCl, if 29.7 g of Br2 and 11.2 g of Cl2 are combined. Determine the amount (in grams) of excess reactant that remains after the reaction is complete.

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  1. 7 July, 13:56
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    There will remain 4.47 grams of Br2

    Explanation:

    Step 1: data given

    Mass of Br2 = 29.7 grams

    Molar mass of Br2 = 159.81 g/mol

    Mass of Cl2 = 11.2 grams

    Molar mass of Cl2 = 70.9 g/mol

    Step 2: The balanced equation

    Br2 + Cl2 → 2BrCl

    Step 3: Calculate moles Br2

    Moles Br2 = 29.7 grams / 159.81 g/mol

    Moles Br2 = 0.186 moles

    Step 4: Calculate moles Cl2

    Moles Cl2 = 11.2 grams / 70.9 g/mol

    Moles Cl2 = 0.158 moles

    Step 5: Calculate limiting reactant

    For 1 mol Br2 we need 1 mol CL2 to produce 2 moles BrCl

    Cl2 is the limiting reactant. It will completely be consumed (0.158 moles)

    Br2 is in excess. There will react 0.158 moles. There will remain 0.186 - 0.158 = 0.028 moles Br2

    Step 6: Calculate moles of Br2 remaining

    Moles Br2 = 0.028 moles * 159.81 g/mol

    Moles Br2 = 4.47 grams

    There will remain 4.47 grams of Br2
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