For the reaction,
$C a O + 2 H C l \to C a C l_{2} + H_{2} O$
1.23 g of CaO is reacted with excess of hydrochloric acid and of $C a C l_{2}$ is formed. What is the per cent yield?

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Solution

The balanced equation is:
$C a O 1 m o l 56 g + 2 H C l \to C a C l_{2} 1 m o l 111 g + H_{2} O$
56 g of $C a O$ produce $C a C l_{2} = 111 g$
1.23 g of $C a O$ produce $C a C l_{2} = \frac{111}{56} \times 1.23 . = 2.43 g$
Thus, Theoretical yield = 2.43 g
Actual yield =1.85 g
Per cent yield $= \frac{1.85}{2.43} \times 100 = 76.1$

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