Question

Solid xenon hexafluoride is prepared by allowing xenon gas and fluorine gas to react.

Xe(g)+3 F2( g)XeF6( s)\mathrm{Xe}(\mathrm{g})+3 \mathrm{~F}_2(\mathrm{~g}) \rightarrow \mathrm{XeF}_6(\mathrm{~s})

How many grams of fluorine are required to produce 10.0 g10.0 \mathrm{~g} of XeF6\mathrm{XeF}_6 ?

Solution

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The mass of fluorine needed for production of XeF6\mathrm{XeF_6} 10.00 g10.00\mathrm{\ g} needs to be calculated.

The reaction of formation of XeF6\mathrm{XeF_6} is:

Xe(g)+3F2(g)XeF6(s)\mathrm{Xe}{(g)}+3\mathrm{F_2}{(g)}\Rightarrow\mathrm{XeF_6}{(s)}

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