Answer
Yield $=88.69\%$
Work Step by Step
Theoretically, the mass of $CaCl_{2}$ that could be obtained by the reaction of $5.57g$ of $Ca(OH)_{2}$ with hydrochloric acid is equal to:
$m(CaCl_{2}) = m(Ca(OH)_{2}) \times \frac{Mr(CaCl_{2})}{Mr(Ca(OH)_{2})}=5.57g \times \frac{111}{74}=8.355g$
Hence, the yield can be calculated by dividing the actual mass of $CaCl_{2}$ obtained by the theoretical mass:
$\alpha = \frac{7.41g}{8.355g}=88.69\%$
