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7 October, 02:13

If 3.491 grams of the precipitate was formed, how many moles of strontium bromide were reacted

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  1. 7 October, 06:05
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    The balanced chemical equation that represents the reaction is as follows:

    SrBr2 (aq) + 2AgNO3 (aq) → Sr (NO3) 2 (aq) + 2AgBr (s)

    From the periodic table:

    mass of silver = 108 grams

    mass of bromine = 80 grams

    molar mass of silver bromide = 108 + 80 = 188 grams

    number of moles = mass / molar mass

    number of moles of produced precipitate = 3.491/188 = 0.018 moles

    From the balanced equation:

    1 mole of strontium bromide produces 2 moles of silver bromide. Therefore, to calculate the number of moles of strontium bromide that produces 0.018 moles of silver bromide, you will just do a cross multiplication as follows:

    amount of strontium bromide = (0.018x1) / 2 = 9.28 x 10^-3 moles
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