Balanced chemical reaction is :
![CaO+2HCl--> CaCl_2+H_2O](https://img.qammunity.org/2021/formulas/chemistry/college/17yx2os7uyzrascypbcy0kmg2n8x6xs9h1.png)
Also 30.2 g of CaO is added to 34.5 g of HCl and 6.35 g of water is formed.
Now , percentage yield is given by :
![\% \ Yield = (Actual\ Yield)/(Theoretical\ Yield )* 100\%](https://img.qammunity.org/2021/formulas/chemistry/college/wmgapxorodw2quco7a3plagjubzom92c2u.png)
Moles of CaO ,
![n_1=(30.2)/(56)=0.54\ moles](https://img.qammunity.org/2021/formulas/chemistry/college/9dcq202kg9fpmttmoyposxocnjxf0algzp.png)
Moles of HCl ,
![n_2=(34.5)/(36.5)=0.95\ moles](https://img.qammunity.org/2021/formulas/chemistry/college/axlji8t4v0uqbhpvqhcmp5ws3srbw7v6vg.png)
Now , 2 moles of HCl react with 1 mole of
.
So , moles of
:
![n=(0.95)/(2)\\\\n=0.475\ moles](https://img.qammunity.org/2021/formulas/chemistry/college/ajc4uffe70n2p7623fhgdjzd25xkf9b534.png)
Mass of water produced :
![m = 0.475 * 18\ g\\\\m=8.55\ g](https://img.qammunity.org/2021/formulas/chemistry/college/hktopau00oi5j56y39shtp0p4knhl4xiqe.png)
But in practical 6.53 g of water is produced .
So ,
![\% \ Yield = (6.35)/(8.55 )* 100\%\\\\\% \ Yield =74.27\%](https://img.qammunity.org/2021/formulas/chemistry/college/kp0ykuuxluovnx35rqy2kxy3qzfvpdsdxy.png)
Therefore, percent yield is 74.27 %.
Hence, this is the required solution.