An excess of $C l_{2}$ reacts with ammonia, the products formed are:

N H_{3}

and

N_{2}

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N C l_{3}

and

H C l

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N C l_{3}

and

N_{2}

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N H_{4} C l

and

N_{2}

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Solution

The correct option is B $N C l_{3}$ and $H C l$
$N H_{3} + 3 C l_{2} (e x c e s s) ⟶ N C l_{3} + 3 H C l$

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