Balance the chemical equation given below and calculate the number of moles of $N H_{4} C l$ formed when 16 moles of $N H_{3}$ react with 6 moles of $C l_{2}$ gas.
$N H_{3} + C l_{2} \to N_{2} + N H_{4} C l$

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Solution

The balanced chemical equation is $8 N H_{3} + 3 C l_{2} \to 6 N H_{4} C l + N_{2}$
According to this reaction $8 moles of N H_{3}$ react with $3 moles of C l_{2}$ to form $6 moles of N H_{4} C l$ .
If we double the amount of both the reactants then the amount of product will also get doubled. That is, when 16 moles of $N H_{3}$ react with 6 moles of $C l_{2}$ then 12 moles of $N H_{4} C l$ will be formed.

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