Answer: The percent yield is, 93.4%
Explanation:
First we have to calculate the moles of Na.
![\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20Na%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20Na%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20Na%7D%7D%3D%5Cfrac%7B41.9g%7D%7B23g%2Fmole%7D%3D1.82moles)
Now we have to calculate the moles of ![Br_2](https://tex.z-dn.net/?f=Br_2)
![{\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles](https://tex.z-dn.net/?f=%7B%5Ctext%7BMoles%20of%7DBr_2%7D%20%3D%20%5Cfrac%7B%5Ctext%7BMass%20of%20%7DBr_2%20%7D%7B%5Ctext%7BMolar%20mass%20of%7D%20Br_2%7D%20%3D%5Cfrac%7B30.3g%7D%7B160g%2Fmole%7D%3D0.189moles)
![{\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles](https://tex.z-dn.net/?f=%7B%5Ctext%7BMoles%20of%20%7D%20NaBr%7D%20%3D%20%5Cfrac%7B%5Ctext%7BMass%20of%20%7D%20NaBr%20%7D%7B%5Ctext%7BMolar%20mass%20of%20%7D%20NaBr%7D%20%3D%5Cfrac%7B36.4g%7D%7B103g%2Fmole%7D%3D0.353moles)
The balanced chemical reaction is,
![2Na(s)+Br_2(g)\rightarrow 2NaBr](https://tex.z-dn.net/?f=2Na%28s%29%2BBr_2%28g%29%5Crightarrow%202NaBr)
As, 1 mole of bromine react with = 2 moles of Sodium
So, 0.189 moles of bromine react with =
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 =
moles of Sodium bromide
Now we have to calculate the percent yield of reaction
![\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%](https://tex.z-dn.net/?f=%5C%25%5Ctext%7B%20yield%7D%3D%5Cfrac%7B%5Ctext%7BActual%20yield%7D%7D%7B%5Ctext%7BTheoretical%20yield%7D%7D%5Ctimes%20100%3D%5Cfrac%7B0.353%20mol%7D%7B0.378%7D%5Ctimes%20100%3D93.4%5C%25)
Therefore, the percent yield is, 93.4%