Respuesta :
Answer: The percent yield is, 93.4%
Explanation:
First we have to calculate the moles of Na.
[tex]\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles[/tex]
Now we have to calculate the moles of [tex]Br_2[/tex]
[tex]{\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles[/tex]
[tex]{\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles[/tex]
The balanced chemical reaction is,
[tex]2Na(s)+Br_2(g)\rightarrow 2NaBr[/tex]
As, 1 mole of bromine react with = 2 moles of Sodium
So, 0.189 moles of bromine react with = [tex]\frac{2}{1}\times 0.189=0.378[/tex] moles of Sodium
Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.
As, 1 mole of bromine give = 2 moles of Sodium bromide
So, 0.189 moles of bromine give = [tex]\frac{2}{1}\times 0.189=0.378[/tex] moles of Sodium bromide
Now we have to calculate the percent yield of reaction
[tex]\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%[/tex]
Therefore, the percent yield is, 93.4%
