Chapter 11, Problem 78A

Interpretation Introduction

Interpretation:

The number of moles of $\text{CsXeF}$ that can be produced from the reaction of 12.5 moles of cesium fluoride with 10.0 moles of xenon hexafluoride needs to be determined.

${\text{CsF}}_{\text{(s)}}{\text{ + XeF}}_{\text{6(s)}}\stackrel{}{\to }{\text{ CsXeF}}_{\text{7(s)}}$

Concept introduction:

The mole concept provides the relation between moles, mass and molar mass. The mathematical expression of mole concept can write as:

$\text{Mole =}\frac{\text{Mass}}{\text{Molar mass}}$

1 mole of any substance contains $\text{6}\text{.022 }\times {\text{ 10}}^{\text{23}}$ number of particles or atoms or molecules.

Answer to Problem 78A

$\text{10}\text{.0 moles CsXeF}$

Explanation of Solution

Given information:

${\text{CsF}}_{\text{(s)}}{\text{ + XeF}}_{\text{6(s)}}\stackrel{}{\to }{\text{ CsXeF}}_{\text{7(s)}}$

Moles of $\text{CsF}$ = 12.5 moles

Moles of ${\text{XeF}}_{\text{6(s)}}$ = 10.0 moles

The given reaction:

${\text{CsF}}_{\text{(s)}}{\text{ + XeF}}_{\text{6(s)}}\stackrel{}{\to }{\text{ CsXeF}}_{\text{7(s)}}$

1 mole of $\text{CsF}$ = 1 mole of ${\text{XeF}}_{\text{6(s)}}$

Determine the limiting reagent:

$\text{10}{\text{.0 moles XeF}}_{\text{6(s)}}\text{×}\frac{\text{1 mole CsF}}{{\text{1 mole XeF}}_{\text{6(s)}}}\text{=10}\text{.0 moles CsF}$

Since $\text{CsF}$ is present in excess so ${\text{XeF}}_{\text{6(s)}}$ must be limiting reagent and will determine the amount of product form.

Calculate moles of $\text{CsXeF}$ with the help of mole ratio:

$\text{10}{\text{.0 moles XeF}}_{\text{6(s)}}\text{ ×}\frac{\text{1 mole CsXeF}}{{\text{1 mole XeF}}_{\text{6(s)}}}\text{= 10}\text{.0 moles CsXeF}$

Conclusion

Mole ratio is used to calculate the mass and moles of product.