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Q.

How many moles of ${CH}_{4}$ will produce $12.0 gm$ ethane according to the reaction:
$\begin{array}{l} {CH}_{4} + {Cl}_{2} \overset{80 %}{\to} {CH}_{3} Cl + HCl \\ 2 {CH}_{3} Cl + 2 Na \overset{50 %}{\to} {CH}_{3} {CH}_{3} + 2 NaCl \end{array}$

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a

$2.0$

b

$0.8$

c

$0.32$

d

$1.0$

answer is A.

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Detailed Solution

$\begin{array}{l} {CH}_{4} + {Cl}_{2} \overset{80 %}{\to} {CH}_{3} Cl + HCl \\ \frac{1}{2} \times \frac{24}{30} \times \frac{1}{8} \frac{1}{2} \times (\frac{24}{30}) \\ 2 {CH}_{3} Cl + 2 Na \overset{50 %}{\to} {CH}_{3} {CH}_{3} + 2 NaCl \\ (\frac{24}{30}) \times 2 \frac{12}{30} \\ n_{{CH}_{4}} = 2 \times \frac{24}{30} \times \frac{5}{4} = 2 moles \end{array}$

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