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11 March, 04:52

It takes 42.25 ml of 0.0500 M Na2S2O3 solution to completely react with the iodine present in a 150.0 ml iodine solution. How many grams of iodine (I2) were initially present in the sample

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  1. 11 March, 05:48
    0
    There were 0.268 grams I2 present

    Explanation:

    Step 1: Data given

    Volume of Na2SO3 = 42.25 mL = 0.04225 L

    Molarity of Na2SO3 = 0.0500 M

    Volume of iodine = 150.0 mL

    Step 2: The balanced equation

    I2 + 2 Na2SO3 → Na2S2O6 + 2 NaI

    Step 3: Calculate moles of Na2SO3

    Moles Na2SO3 = molarity * volume

    Moles Na2SO3 = 0.0500 M * 0.04225 L

    Moles Na2SO3 = 0.0021125 moles

    Step 4: Calculate moles I2

    For 1 mol I2 we need 2 moles Na2SO3

    For 0.0021125 moles Na2SO3 we have 0.00105625

    Step 5: Calculate mass of I2

    Mass I2 = moles * molar mass

    Mass I2 = 0.00105625 * 253.8 g/mol

    Mass I2 = 0.268 grams

    There were 0.268 grams I2 present
  2. 11 March, 08:05
    0
    Answer: I lived in street when I was twelve years old
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