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# If you started with 5.0 L of a 0.25 M solution of xenon trioxide, and you isolated 24 g of oxygen, what is your percent yield?

Balanced equation: 2 XeO3 ----> 2 Xe + O3

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- Roger the MoleLv 78 months ago
The equation shown is not balanced. I think it's supposed to be:

2 XeO3 → 2 Xe + 3 O2

(5.0 L) x (0.25 mol XeO3 / L) x (3 mol O2 / 2 mol XeO3) x (31.99886 g O2/mol) = 59.9978 g O2 in theory

(24 g) / (59.9978 g) = 0.40001 = 40.% yield O2

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