Question

If 30.2 g of CaO is added to 34.5 g of HCl and 6.35 g of water is formed, what

Answer 1

Balanced chemical reaction is :

$CaO+2HCl--> CaCl_2+H_2O$

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 :

$\% \ Yield = \dfrac{Actual\ Yield}{Theoretical\ Yield }\times 100\%$

Moles of CaO , $n_1=\dfrac{30.2}{56}=0.54\ moles$

Moles of HCl , $n_2=\dfrac{34.5}{36.5}=0.95\ moles$

Now , 2 moles of HCl react with 1 mole of $H_2O$ .

So , moles of $H_2O$ :

$n=\dfrac{0.95}{2}\\\\n=0.475\ moles$

Mass of water produced :

$m = 0.475 \times 18\ g\\\\m=8.55\ g$

But in practical 6.53 g of water is produced .

So ,

$\% \ Yield = \dfrac{6.35}{8.55 }\times 100\%\\\\\% \ Yield =74.27\%$

Therefore, percent yield is 74.27 %.

Hence, this is the required solution.