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

This is about as straight forward as you can get in a limiting reactants problem.

Each mol of cesium xenon heptafluoride will require 1 mol each of the CsF and xenon hexafluoride.

From this relationship, you can see that the limiting reactant is the CsF. Now convert mols of CsF to mols of CsXeF_7:
\[11molCsF*\frac{1molCsXeF_{7}}{1molCsF}=11molCsXeF_{7}\]
11mols of CsXeF_7 can be produced.