Answer :
(a) The mass of
produced is, 15.2 grams.
(b) The percent yield of the reaction is, 72.5 %
Explanation :
Part (a) :
Given,
Mass of
= 85.1 g
Molar mass of
= 27 g/mol
First we have to calculate the moles of ![Al](https://tex.z-dn.net/?f=Al)
![\text{Moles of }Al=\frac{\text{Given mass }Al}{\text{Molar mass }Al}=\frac{85.1g}{27g/mol}=3.15mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DAl%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%20%7DAl%7D%7B%5Ctext%7BMolar%20mass%20%7DAl%7D%3D%5Cfrac%7B85.1g%7D%7B27g%2Fmol%7D%3D3.15mol)
Now we have to calculate the moles of ![Al_2O_3](https://tex.z-dn.net/?f=Al_2O_3)
The balanced chemical equation is:
![4Al+3O_2\rightarrow 2Al_2O_3](https://tex.z-dn.net/?f=4Al%2B3O_2%5Crightarrow%202Al_2O_3)
From the reaction, we conclude that
As, 4 moles of
react to give 2 moles of ![Al_2O_3](https://tex.z-dn.net/?f=Al_2O_3)
So, 3.15 moles of
react to give
mole of ![Al_2O_3](https://tex.z-dn.net/?f=Al_2O_3)
Now we have to calculate the mass of ![Al_2O_3](https://tex.z-dn.net/?f=Al_2O_3)
![\text{ Mass of }Al_2O_3=\text{ Moles of }Al_2O_3\times \text{ Molar mass of }Al_2O_3](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DAl_2O_3%3D%5Ctext%7B%20Moles%20of%20%7DAl_2O_3%5Ctimes%20%5Ctext%7B%20Molar%20mass%20of%20%7DAl_2O_3)
Molar mass of
= 102 g/mole
![\text{ Mass of }Al_2O_3=(1.58moles)\times (102g/mole)=161.2g](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DAl_2O_3%3D%281.58moles%29%5Ctimes%20%28102g%2Fmole%29%3D161.2g)
Therefore, the mass of
produced is, 161.2 grams.
Part (b) :
Now we have to calculate the percent yield of the reaction.
![\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100](https://tex.z-dn.net/?f=%5Ctext%7BPercent%20yield%7D%3D%5Cfrac%7B%5Ctext%7BExperimental%20yield%7D%7D%7B%5Ctext%7BTheoretical%20yield%7D%7D%5Ctimes%20100)
Experimental yield = 116.9 g
Theoretical yield = 161.2 g
Now put all the given values in this formula, we get:
![\text{Percent yield}=\frac{116.9g}{161.2g}\times 100=72.5\%](https://tex.z-dn.net/?f=%5Ctext%7BPercent%20yield%7D%3D%5Cfrac%7B116.9g%7D%7B161.2g%7D%5Ctimes%20100%3D72.5%5C%25)
Therefore, the percent yield of the reaction is, 72.5 %