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Anonymous asked in Science & MathematicsChemistry · 2 years ago

S8(s) + 24 F2(g) → 8 SF6(g) What is the percent yield if 18.3 g SF6 is isolated from the reaction of 30.0 g F2 and 10.0 g sulfur?

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  • 2 years ago
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    (30.0 g F2) / (37.9968 g F2/mol) = 0.78954 mol F2

    (10.0 g S8) / (256.520 g S8/mol) = 0.038983 mol S8

    0.78954 mole of F2 would react completely with 0.78954 x (1/24) = 0.0328975 mole S8, but there is more S8 present than that, so S8 is in excess and F2 is the limiting reactant.

    (0.78954 mol F2) x (8 mol SF6 / 24 mol F2) x (146.0554 g SF6/mol) = 38.439 g SF6 in theory

    (18.3 g SF6) / (38.439 g SF6) = 0.476 = 47.6% yield SF6

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