Question

Observe the following chemical reaction,
\[{\text{HCl (g) + N}}{{\text{H}}_{\text{3}}}{\text{(g) }} \to {\text{ N}}{{\text{H}}_{\text{4}}}{\text{Cl (s) }}\]
If 3.0 moles of HCl gas and 3 moles of \[{\text{N}}{{\text{H}}_{\text{3}}}\]gas each measured at \[{\text{20}}^\circ {\text{C}}\]and 1.0 atmosphere pressure, are allowed to react completely according the equation above, the final mixture will contain
A.3 moles of solid \[{\text{ N}}{{\text{H}}_{\text{4}}}{\text{Cl }}\]only
B.5 moles of solid \[{\text{ N}}{{\text{H}}_{\text{4}}}{\text{Cl }}\]only
C.3 moles of solid \[{\text{ N}}{{\text{H}}_{\text{4}}}{\text{Cl }}\]+ 2 moles of \[{\text{N}}{{\text{H}}_{\text{3}}}\]gas
D.3 moles of solid \[{\text{ N}}{{\text{H}}_{\text{4}}}{\text{Cl }}\]+ 2 moles of \[{\text{HCl}}\]gas
D.2 moles of \[{\text{HCl}}\]gas, 4 moles of \[{\text{N}}{{\text{H}}_{\text{3}}}\]gas and 1 mole of solid \[{\text{ N}}{{\text{H}}_{\text{4}}}{\text{Cl }}\]

Answer 1

Hint: Using the given balanced reaction determine the mole ratio of \[{\text{HCl}}\]:\[{\text{N}}{{\text{H}}_{\text{3}}}\]:\[{\text{ N}}{{\text{H}}_{\text{4}}}{\text{Cl }}\]. Using the mole ratio determines the moles of product formed. Also, determine the unreacted moles of the reactant if they remain in the mixture after completion of the reaction.

Complete step by step answer:
The reaction given to us is:
\[{\text{HCl (g) + N}}{{\text{H}}_{\text{3}}}{\text{(g) }} \to {\text{ N}}{{\text{H}}_{\text{4}}}{\text{Cl (s) }}\]
From this balance reaction, we can say that 1 mole of \[{\text{HCl}}\] reacts with 1 mole of \[{\text{N}}{{\text{H}}_{\text{3}}}\]and produce 1 mole of\[{\text{ N}}{{\text{H}}_{\text{4}}}{\text{Cl }}\].
So, the mole ratio of \[{\text{HCl}}\]:\[{\text{N}}{{\text{H}}_{\text{3}}}\]:\[{\text{ N}}{{\text{H}}_{\text{4}}}{\text{Cl }}\] is 1:1:1.
Using the mole ratio of reactants and product calculate the moles of product form as follows:
We have given 3.0 moles of HCl gas and 3 moles of \[{\text{N}}{{\text{H}}_{\text{3}}}\] gas.
As mole ratio of \[{\text{HCl}}\]:\[{\text{N}}{{\text{H}}_{\text{3}}}\]:\[{\text{ N}}{{\text{H}}_{\text{4}}}{\text{Cl }}\] is 1:1:1 and moles of HCl gas and moles of \[{\text{N}}{{\text{H}}_{\text{3}}}\] gas given is also equal so 3.0 moles of HCl gas will react completely with 3 moles of \[{\text{N}}{{\text{H}}_{\text{3}}}\]gas and will give 3.0 moles of solid\[{\text{ N}}{{\text{H}}_{\text{4}}}{\text{Cl }}\].
As both the reactants \[{\text{HCl}}\] gas and \[{\text{N}}{{\text{H}}_{\text{3}}}\]gas react completely with each other there will be no reactant after completion of the reaction.
So, after completion of the reaction, final mixture will contain only 3.0 moles of solid\[{\text{ N}}{{\text{H}}_{\text{4}}}{\text{Cl }}\].

Thus, the correct option is (A) 3 moles of solid \[{\text{ N}}{{\text{H}}_{\text{4}}}{\text{Cl }}\] only.

Note: Always use the balance reaction to determine the mole ratio of reactant and product. Moles of product form and moles of reactants remain unreacted depending on the mole ratio of reactant and product and given moles of reactants.