Question

One of the few xenon compounds that form is cesium xenon heptafluoride (CsXeF7). How many moles of CsXeF7 can be produced from the reaction of 11.0 mol cesium fluoride with 11.5 mol xenon hexafluoride?
CsF(s) + XeF6(s) CsXeF7(s) ...?

Answer 1

The balanced chemical equation is written as:

CsF(s) + XeF6(s) ------> CsXeF7(s)

We are given the amount of cesium fluoride and xenon hexafluoride used for the reaction. We need to determine first the limiting reactant to proceed with the calculation. From the equation and the amounts, we can say that the limiting reactant would be cesium fluoride.  We calculate as follows:

11.0 mol CsF ( 1 mol CsXeF7 / 1 mol CsF ) = 11.0 mol CsXeF7

Answer 2

Answer: 11 moles

Explanation:

$CsF(s)+XeF_6(s)\rightarrow CsXeF_7(s)$

As can be seen from the balanced chemical reaction,

1 mole of CsF reacts with 1 mole of $XeF_6$

Thus 11 moles of CsF react with 11 mole of $XeF_6$

Thus CsF is the limiting reagent as it limits the formation of products and $XeF_6$ is the excess reagent as (11.5-11)=0.5 moles are in excess.

As  1 mole of CsF produces= 1 mole of $CsXeF_7$

11 moles of CsF will produce=$\frac{1}{1}\times 11=11 moles$ of $CsXeF_7$