$11.2 g$ of quicklime ( $CaO$ , molar mass $56 g mo l^{- 1}$ ) is slaked completely with water. The resulting hydroxide is treated with excess chlorine at 313 K to form bleaching powder $CaOCl X_{2}$ (molar mass $127 g mo l^{- 1}$ ) as per $Ca (OH) X_{2} + Cl X_{2} CaOCl X_{2} + H X_{2} O$ , in $80%$ yield. The mass of $CaOCl X_{2}$ obtained is:

Answer: D

CaO + H X_{2} O Ca (OH) X_{2}

, and 11.2 g is

11.2/56 = 0.2

mol, so 0.2 mol

Ca (OH) X_{2}

forms. The theoretical mass of

CaOCl X_{2}

0.2 \times 127 = 25.40

g. At 80% yield it is 25.40

\times 0.8 = 20.32