$C l_{2}$ required in this reaction is :

0.1221

kmol

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8.666

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4.333

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0.1221

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Solution

The correct options are
A $8.666$ kg
B $0.1221$ kmol
The reaction is as follows:
$2 T i O_{2} + 3 C + 4 C l_{2} 70 % - - \to 2 T i C l_{4} + C O_{2} + 2 C O$
$7500$ g
$\frac{7500}{80} = 93.75$ mol
So, $C l_{2}$ requires $= 93.75 \times 2 = 187.5 \times 0.7 \times 0.93 = 122.1$ mol $= 8.66$ kg.

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