Question

A fluorine disposal plant was constructed to carry out the reactions:
$F_2+2NaOH\to \frac{1}{2}{O}_2+2NaF+H_{2}O$
$2NaF+CaO+H_{2}O\to Ca{F}_2+2NaOH$
As the plant operates, excess lime was added to bring about complete precipitation of the fluoride as $Ca{F}_2$. Over a period of operation, 1900 kg of fluorine was fed into a plant and 10,000 kg of lime was required. What was the percentage utillzation of lime? [Lime: CaO]
%

Answer 1

The given balanced reactions are:
$F_2+2NaOH\to \frac{1}{2}{O}_2+2NaF+H_{2}O$
$2NaF+CaO+H_{2}O\to Ca{F}_2+2NaOH$

$F_2$ added = 1900 kg $=1900\times {10}^{3}g$
$=\frac{19\times {10}^3}{38}\text{moles}=\frac{1\times {10}^2}{2}\text{moles}$
(molar mass of $F_2=38g/mol)$

Moles of $NaF$ formed = 2 (moles of $F_2)$
$=2(0.5\times {10}^5\text{moles})=1\times {10}^5\text{moles}$

Moles of $CaO$, required $=\frac{1}{2}(1\times {10}^5\text{moles})$
Mass used $=0.5\times {10}^5\times 56$
$=28\times {10}^{5}g=28\times {10}^{2}kg$
(molar mass of $CaO$ = 56 g/mol)

Feed amount of lime = 10,000 kg
$\because$ % Utilization of $CaO=\frac{2800}{10,000}\times 100=28\%$