Final answer:
Starting with 14.0 mol of both cesium fluoride and xenon hexafluoride, you can produce 14.0 moles of cesium xenon heptafluoride due to the 1:1 molar ratio of the reactants according to the balanced chemical equation.
Step-by-step explanation:
The student's question asks how many moles of cesium xenon heptafluoride (CsXeF₇) can be produced from the reaction of 14.0 mol cesium fluoride with 14.0 mol xenon hexafluoride according to the equation CsF(s) + XeF₆(s) → CsXeF₇(s).
To answer this question, you need to look at the stoichiometry of the balanced equation. The balanced equation shows a 1:1 molar ratio between CsF and XeF₆. Because both reactants are present in equal amounts, neither one is in excess, and they will combine to form CsXeF₇ in a 1:1 molar ratio.
Therefore, when you start with 14.0 mol of CsF and 14.0 mol of XeF₆, all of the reactants will be completely consumed to produce 14.0 moles of CsXeF₇, assuming the reaction goes to completion without any side reactions or losses.