Question

10 g of hydrogen and 28 g nitrogen are allowed to react and the compound produced (assume complete forward reaction) is dissolved in one litre of water to produce a basic solution. The molarity of solution is:

• A. $2M$
• B. $1M$
• C. $5M$
• D. $3M$

Answer 1

The correct option is A $2M$

Nitrogen reacts with hydrogen to produce ammonia according to the following reaction:
$N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$
Here, we are assuming the gases react only in the forward reaction and there is nealy no decomposition of ammonia to give back $N_2\left(g\right)$ & $H_2\left(g\right)$.
So 1 mol of $N_2$ reacts 3 mol of $H_2$ to give 2 mol of $N{H}_3$. Here, we are given $\frac{10}{2}=5mol$ of $H_2$ and $\frac{28}{28}=1mol$ of $N_2$. So $N_2$ is the limiting reagent here and accordingly $2moles$ of $N{H}_3$ will be produced. Thus, the required molarity $=\frac{\text{Moles of}N{H}_3}{\text{Volume of soln in L}}=\frac{2}{1}=2M$