Question

Chlorine reacts with ammonia and gives nitrogen. The ratio of chlorine to ammonia in this reaction is:

Answer 1

Hint: Write chlorine molecule reacting with ammonia to give nitrogen. Along with nitrogen there is one more product i.e. Hydrochloric acid. Now balance the atoms on the reactants side as well as on the products side in order to form a balanced reaction. Now take the ratio of stoichiometric coefficient of chlorine and ammonia to arrive at the answer.

Complete step-by-step answer:
Chlorine is a chemical element having the symbol "Cl" and atomic number 17. It is the second-lightest of the halogens, appearing after fluorine in the periodic table and its characteristics are between fluorine and bromine.
Chlorine exists as yellow-green gas at room temperature. It is a highly reactive element as well as a strong oxidizing agent. Among the halogen elements, it has the highest electron affinity and the third-highest electronegativity on the Pauling scale after fluorine and oxygen.
Ammonia is a chemical compound consisting of nitrogen and hydrogen. It is a stable binary hydride, and the simplest pnictogen hydride (15th group elements are called pnictogens). Ammonia exists a colorless gas with a characteristic pungent smell at room temperature.
Ammonia is mainly used either directly or indirectly, as a building block for the synthesis of many pharmaceutical products. Along with that it is used in many commercial cleaning products.
We will now write the reaction between ammonia and chlorine to obtain the products.
$\text{3C}{{\text{l}}_{\text{2}}}\text{ + 8N}{{\text{H}}_{\text{3}}}\text{ }\to \text{ }{{\text{N}}_{\text{2}}}\text{ + 6N}{{\text{H}}_{4}}\text{Cl}$
In the balanced reaction we see that 3 moles of chlorine gas reacts with 8 moles of ammonia gas to release nitrogen gas as product.

The ratio of chlorine to ammonia thus becomes equal to 3:8. Therefore, the correct answer is option (C).

Note: When ammonia reacts with chlorine in excess quantity the products formed are $\text{NC}{{\text{l}}_{\text{3}}}$ and $\text{HCl}$.
$\text{N}{{\text{H}}_{\text{3}}}\text{ + 3C}{{\text{l}}_{\text{2(excess)}}}\text{ }\to \text{ NC}{{\text{l}}_{\text{3}}}\text{ + 3HCl}$