How many grams of xenon are required to form 20.0 g of xenon hexafluoride? What is the balanced equation for this reaction? I am unsure of how to find the mass without the right equation. Please...

How many grams of xenon are required to form 20.0 g of xenon hexafluoride?

What is the balanced equation for this reaction? I am unsure of how to find the mass without the right equation.

Please explain.

Asked on by takenotes

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jerichorayel | College Teacher | (Level 2) Senior Educator

Posted on

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. 

Sources:

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