Answer: The percent yield for the [tex]NaBr[/tex] is, 86.7 %
Explanation : Given,
Moles of [tex]FeBr_3[/tex] = 2.36 mol
Moles of [tex]NaBr[/tex] = 6.14 mol
First we have to calculate the moles of [tex]NaBr[/tex]
The balanced chemical equation is:
[tex]2FeBr_3+3Na_2S\rightarrow Fe_2S_3+6NaBr[/tex]
From the reaction, we conclude that
As, 2 moles of [tex]FeBr_3[/tex] react to give 6 moles of [tex]NaBr[/tex]
So, 2.36 moles of [tex]FeBr_3[/tex] react to give [tex]\frac{6}{2}\times 2.36=7.08[/tex] mole of [tex]NaBr[/tex]
Now we have to calculate the percent yield for the [tex]NaBr[/tex].
[tex]\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield = 6.14 moles
Theoretical yield = 7.08 moles
Now put all the given values in this formula, we get:
[tex]\text{Percent yield}=\frac{6.14mol}{7.08mol}\times 100=86.7\%[/tex]
Therefore, the percent yield for the [tex]NaBr[/tex] is, 86.7 %
