Balanced chemical reaction is :
[tex]CaO+2HCl--> CaCl_2+H_2O[/tex]
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 :
[tex]\% \ Yield = \dfrac{Actual\ Yield}{Theoretical\ Yield }\times 100\%[/tex]
Moles of CaO , [tex]n_1=\dfrac{30.2}{56}=0.54\ moles[/tex]
Moles of HCl , [tex]n_2=\dfrac{34.5}{36.5}=0.95\ moles[/tex]
Now , 2 moles of HCl react with 1 mole of [tex]H_2O[/tex] .
So , moles of [tex]H_2O[/tex] :
[tex]n=\dfrac{0.95}{2}\\\\n=0.475\ moles[/tex]
Mass of water produced :
[tex]m = 0.475 \times 18\ g\\\\m=8.55\ g[/tex]
But in practical 6.53 g of water is produced .
So ,
[tex]\% \ Yield = \dfrac{6.35}{8.55 }\times 100\%\\\\\% \ Yield =74.27\%[/tex]
Therefore, percent yield is 74.27 %.
Hence, this is the required solution.
