The reaction proceeds as :
$$
\begin{gathered}
\mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3 \\
28 \mathrm{~g} 6 \mathrm{~g} \quad 34 \mathrm{~g}
\end{gathered}
$$
$\because 6 \mathrm{~g} \mathrm{H}_2$ reacts with $28 \mathrm{~g} \mathrm{~N}_2$.
$$
\therefore 10 \mathrm{~g} \mathrm{H}_2 \text { reacts with } \frac{28}{6} \times 10=46.67 \mathrm{~g} \mathrm{~N}_2
$$
i.e. $\mathrm{H}_2$ is limiting reagent and $\mathrm{N}_2$ is in excess.
Now,
$$
\begin{aligned}
& \because 6 \mathrm{~g} \mathrm{H}_2 \text { produces } 34 \mathrm{~g} \mathrm{NH}_3 . \\
& \therefore 10 \mathrm{~g} \mathrm{H}_2 \text { produces }\left(\frac{34}{6} \times 10\right) \mathrm{gNH}_3 .
\end{aligned}
$$
Therefore, we can calculate number of moles of $\mathrm{NH}_3$ as
$$
\begin{aligned}
n_{\mathrm{NH}_3} & =\frac{\text { Produced weight }}{\text { Molecular weight }} \\
& =\frac{\left(\frac{34}{6} \times 10\right)}{17}=\frac{34}{6} \times \frac{10}{17}=\frac{10}{3}=3.33
\end{aligned}
$$