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How many grams of xenon are required to form 20.0 g of xenon hexafluoride? What is the...
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The balanced equation for the reaction can be written as:
`Xe + 3 F_2 -> XeF_6`
The balanced equation shows that one mole of xenon and three moles of fluorine gas is needed to form one mole of xenon hexafluoride. Fluorine is a diatomic molecule which normally exists as `F_2` .We can find the amount of xenon needed using stoichiometry.
`20.0 g XeF_6 * (1 mol e XeF_6)/(245.28 grams XeF_6) * (1 mol e Xe)/(1mol e XeF_6) * (131.293 grams Xe)/(1mol e Xe)`
= 10.7 grams Xe is required to form the xenon hexafluoride compound.
Posted by jerichorayel on May 10, 2013 at 12:04 AM (Answer #1)
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