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14 October, 15:28

How many bromines atoms are present in 33.3 g of ch2br2?

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  1. 14 October, 15:29
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    37.8 g CH2Br2 X (1 mol CH2Br2 / 173.83 g) = 4.60X10^-3 mol CH2Br2

    4.60X10^-3 mol CH2Br2 X (2 mol Br / 1 mol CH2Br2) X 6.02X10^23 atoms/mol = 5.54X10^21 bromine atoms.

    I think this is the answer.
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