Question

A sample of 6.81 moles of cesium fluoride is

reacted with 8.23 moles of xenon hexafluoride
to produce cesium xenon heptafluoride. Using
the balanced equation below, calculate the
maximum amount in moles of cesium xenon
heptafluoride that can be produced.
CsF+XeF6- CsXeF7

Answer 1

6.91 moles cesium xenon heptafluoride

Step-by-step explanation:

The balanced equation

CsF + XeF6 = CsXeF7

tells us that we should expect 1 mole of cesium xenonheptafluoride (CsXeF7) for every 1 mole of cesium fluoride consumed, a molar ratio of 1:1. It also says we will use 1 mole of xenon hexafluoride, XeF6, for each mole of cesium fluoride.

The 6.91 moles of CsF will therefore require 6.91 moles of XeF6 to completely react. We are in luck, since there are 8.23 moles of XeF6, more than enough to react with all the CsF. In fact, there should be (8.23 moles - 6.91 moles =) 1.32 moles of XeF6 left over, unreacted. CsF is the "limiting reagent:" the reaction will stop once it is consumed. We can assume that all of the CsF will react, so that will result in the same number of moles of CsXeF7 produced, since the molar ratio of the two is 1:1 reactant/product.