What was the percentage utilisation of lime ( $C a O$ )?

29.5 %

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14.75 %

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59.0 %

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60.0 %

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Solution

The correct option is A $29.5 %$
$F_{2} + 2 N a O H ⟶ \frac{1}{2} O_{2} + 2 N a F + H_{2} O$
$2 N a F + C a O + H_{2} O ⟶ C a F_{2} + 2 N a O H$
According to process,
$F_{2}$ $C a O$
$1$ mole $1$ mole
$\frac{2000}{38}$ $\frac{10, 000}{56}$
$= 52.63$ kmol $= 178.56$ kmol
So, $%$ of $C a O$ used $= \frac{52.63 \times 100}{178.56} = 29.5 %$ .

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