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19 October, 06:57

How many grams of fluorine are required to produce 10.0 g of XeF6?

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  1. 19 October, 07:13
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    Moles XeF6 = 10.0 g / 245.28 g/mol=0.0408

    the ratio between F2 and XeF6 is 3 : 1

    moles F2 required = 3 x 0.0408=0.122

    mass F2 = 0.122 mol x 37.9968 g/mol=4.64 g
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