\(\begin{array}{l}
a)\\
nNaOH = n{H_2}S{O_4} = \dfrac{{4,9}}{{98}} = 0,05\,mol\\
mNaOH = 0,05 \times 40 = 2g\\
b)\\
n{N_2} = n{O_2} = \dfrac{{1,6}}{{32}} = 0,05\,mol\\
m{N_2} = 0,05 \times 28 = 1,4g\\
c)\\
nN{H_3} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
AN{H_3} = 0,25 \times 6 \times {10^{23}} = 1,5 \times 6 \times {10^{23}}\\
d)\\
nFe = nC{O_2} = \dfrac{{2,2}}{{44}} = 0,05\,mol
\end{array}\)