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 10.5 mol xenon hexafluoride?
CsF(s) + XeF6(s) CsXeF7(s)
mol

Answer 1

Answer:

10.5 mole

Explanation:

Data given:

no. of mole xenon hexafluoride = 10.5 moles

no. of mole cesium fluoride  = 11 moles

moles of CsXeF₇ = ?

Reaction Given:

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

Solution:

First we will look for the reaction

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

if we look for the mole ratio of CsF to XeF₆

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

               1 mole       1 mole

It means there is 1:1 that is one mole of  CsF combine 1 mole XeF₆ to form CsXeF₇. So if we look at the given data cesium fluoride is in excess than XeF₆

Excess amount of CsF = 11.0 mole - 10.5 mole = 0.5 mole

So XeF₆ is limiting reactant and formation of product depends on the amount of XeF₆.

So again look at the reaction on the base of limiting reactant

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

                                   1 mole                1 mole

it means 1 mole XeF₆ produces 1 mole of CsXeF₇, then how many moles of CsXeF₇ will be produce from 10.5 moles of XeF₆

Apply unity formula

                  1 mole XeF₆ ≅ 1 mole of CsXeF₇

                  10.5 mole XeF₆ ≅ X mole of CsXeF₇

Do cross multiplication

             mole of CsXeF₇ =  10.5 mole x 1 mole / 1 mole

             mole of CsXeF₇ =  10.5 mole

So,

10.5 mole of CsXeF₇ will be produces.