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

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Solution

$2 T i O_{2} + 3 C + 4 C l_{2} 70 % - - \to 2 T i C l_{4} + C O_{2} + 2 C O$
$7500$ g
$\frac{7500}{80} = 93.75$ mole
So, $C l_{2}$ requires $= 93.75 \times 2 = 187.5 \times 0.7 \times 0.93 = 122.1$ mole $= 8.66 k g$ .

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