In Haber's process 50.0 g of $N_2$ (g) and 10.0g of $H_2$ (g) are mixed to produce $N{H}_3$ (g). What the number of moles of $N{H}_3$ (g) formed?

Number of moles of nitrogen $=weight/molecularweight=50/28=1.785$ moles         

Number of moles of hydrogen $=weight/molecularweight=10/2=5$ moles   

The balanced equation is 

$$\begin{gathered} N_2 + 3H_2 \to 2N{H}_3 \\ 1 3 2moles \\ given, 1.785 5 ? \end{gathered}$$

Here hydrogen is the limiting reagent. As it is known that three mole of       

hydrogen to form two moles of ammonia.

Then five moles of hydrogen will produce $2/3\times 5=3.33$

Hence the correct option is (A).