Correct Answer -(b)\(2g\,H_2\)
7g of \(N_2=\frac14 \text{ mol of } N_2\)
2g of \(H_2=1 \text{ mol of } H_2\)
16g of \(NO_2=0.35\text{ mol of } NO_2\)
16g of \(O_2=0.5 \text{ mol of } O_2\)
\(\therefore\) 2g of \(H_2\) has maximum number of molecules.