Question

$28\text{ g}$ of ${{N}_{2}}$ and $6g$ of ${{H}_{2}}$ were mixed. At equilibrium, $17g$ g of $N{{H}_{3}}$ was produced. Then, the weights of ${{N}_{2}}$ and ${{H}_{2}}$ at equilibrium are respectively
(a) $11\text{ g , 0g}$
(b) $1\text{ g , 3g}$
(c) $14\text{ g , 3g}$
(d) $11\text{ g , 3g}$

Answer 1

Hint: First of we should know the number of the nitrogen and hydrogen at the equilibrium and then by using them, we can easily calculate the weight of the nitrogen and hydrogen at equilibrium by using the formula as; $mole=\dfrac{given\text{ }weight}{molecular\text{ }weight}$. Now solve it.

Complete step by step answer:
- When nitrogen reacts with the hydrogen , it results in the formation of the ammonia. The reaction of the formation of the ammonia occurs as:
${{N}_{2}}+3{{H}_{2}}\to 2N{{H}_{3}}$
Initially, the number of moles present are ; $\begin{align}
 & {{N}_{2}}+3{{H}_{2}}\to 2N{{H}_{3}} \\

\end{align}$
At equilibrium, the number of moles present are ; $\begin{align}
 & {{N}_{2}}+3{{H}_{2}}\to 2N{{H}_{3}} \\

 & \text{1-x 3-3x 2x} \\
\end{align}$
- Now, from this we can calculate the number of moles of the nitrogen and hydrogen at equilibrium as;
First of we will find the value of x by using the number of moles of ammonia as;
$\begin{align}
 & 2x=1 \\
 & x=\dfrac{1}{2} \\
\end{align}$
- Now, from this, we can calculate the number of moles of nitrogen present at equilibrium is as:
Moles of nitrogen:- $1-\dfrac{1}{2}=\dfrac{2-1}{2}=\dfrac{1}{2}$
Similarly, we can calculate the number of moles hydrogen at the equilibrium as;
Moles of hydrogen:- $3-\dfrac{3}{2}=\dfrac{6-3}{2}=\dfrac{3}{2}$
- Now, we know the number of moles of the nitrogen and hydrogen and by using the mole formula we can easily calculate, the given weight of the nitrogen and the hydrogen.
$\begin{align}
 & mole=\dfrac{given\text{ }weight}{molecular\text{ }weight} \\
 & given\text{ }weight=mole\times molecular\text{ }weight-------(1) \\
\end{align}$
We know that;
The molecular weight of nitrogen=$28$
And the molecular weight of hydrogen=$2$
- Now using the equation (1), we can calculate the given mass as;
$\begin{align}
 & given\text{ }weight\text{ }of\text{ }{{\text{N}}_{2}}=\dfrac{1}{2}\times 28 \\
 & \text{ =14g} \\
\end{align}$
$\begin{align}
 & given\text{ }weight\text{ }of\text{ }{{\text{H}}_{2}}=\dfrac{3}{2}\times 2 \\
 & \text{ =3g} \\
\end{align}$
Thus, when the $28\text{ g}$ of ${{N}_{2}}$ and $6g$ of ${{H}_{2}}$ were mixed together, $17g$ g of $N{{H}_{3}}$ was produced. Then, the weights of ${{N}_{2}}$ and ${{H}_{2}}$ at equilibrium are $14\text{ g and 3g}$.
The correct option is option “C” .

Note: Always remember that the no. of moles of each gas at the standard conditions of the temperature and pressure, contains Avogadro number of particles and occupies the volume of $22.4\text{ liters}$.