Answer : The mass of product produced if the reaction occurred with a 70.5 percent yield will be, 20.67 grams.
Explanation : Given,
Mass of P = 25 g
Mass of
= 25 g
Molar mass of P = 30.97 g/mole
Molar mass of
= 71 g/mole
Molar mass of
= 208.24 g/mole
First we have to calculate the moles of
and
.
![\text{Moles of }P=\frac{\text{Mass of }P}{\text{Molar mass of }P}=\frac{25g}{30.97g/mole}=0.807moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DP%3D%5Cfrac%7B%5Ctext%7BMass%20of%20%7DP%7D%7B%5Ctext%7BMolar%20mass%20of%20%7DP%7D%3D%5Cfrac%7B25g%7D%7B30.97g%2Fmole%7D%3D0.807moles)
![\text{Moles of }Cl_2=\frac{\text{Mass of }Cl_2}{\text{Molar mass of }Cl_2}=\frac{25g}{71g/mole}=0.352moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DCl_2%3D%5Cfrac%7B%5Ctext%7BMass%20of%20%7DCl_2%7D%7B%5Ctext%7BMolar%20mass%20of%20%7DCl_2%7D%3D%5Cfrac%7B25g%7D%7B71g%2Fmole%7D%3D0.352moles)
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
![2P+5Cl_2\rightarrow 2PCl_5](https://tex.z-dn.net/?f=2P%2B5Cl_2%5Crightarrow%202PCl_5)
From the balanced reaction we conclude that
As, 5 moles of
react with 2 moles of ![P](https://tex.z-dn.net/?f=P)
So, 0.352 moles of
react with
moles of ![P](https://tex.z-dn.net/?f=P)
That means, in the given balanced reaction,
is a limiting reagent and it limits the formation of products and
is an excess reagent because the given moles are more than the required moles.
Now we have to calculate the moles of
.
As, 5 moles of
react with 2 moles of ![PCl_5](https://tex.z-dn.net/?f=PCl_5)
So, 0.352 moles of
react with
moles of ![PCl_5](https://tex.z-dn.net/?f=PCl_5)
Now we have to calculate the mass of
.
![\text{Mass of }PCl_5=\text{Moles of }PCl_5\times \text{Molar mass of }PCl_5](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20%7DPCl_5%3D%5Ctext%7BMoles%20of%20%7DPCl_5%5Ctimes%20%5Ctext%7BMolar%20mass%20of%20%7DPCl_5)
![\text{Mass of }PCl_5=(0.1408mole)\times (208.24g/mole)=29.32g](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20%7DPCl_5%3D%280.1408mole%29%5Ctimes%20%28208.24g%2Fmole%29%3D29.32g)
Now we have to calculate the mass of product produced (actual yield).
![\%\text{ yield of }PCl_5=\frac{\text{Actual yield of }PCl_5}{\text{Theoretical yield of }PCl_5}\times 100](https://tex.z-dn.net/?f=%5C%25%5Ctext%7B%20yield%20of%20%7DPCl_5%3D%5Cfrac%7B%5Ctext%7BActual%20yield%20of%20%7DPCl_5%7D%7B%5Ctext%7BTheoretical%20yield%20of%20%7DPCl_5%7D%5Ctimes%20100)
![70.5=\frac{\text{Actual yield of }PCl_5}{29.32g}\times 100](https://tex.z-dn.net/?f=70.5%3D%5Cfrac%7B%5Ctext%7BActual%20yield%20of%20%7DPCl_5%7D%7B29.32g%7D%5Ctimes%20100)
![\text{Actual yield of }PCl_5=20.67g](https://tex.z-dn.net/?f=%5Ctext%7BActual%20yield%20of%20%7DPCl_5%3D20.67g)
Therefore, the mass of product produced if the reaction occurred with a 70.5 percent yield will be, 20.67 grams.