d
since, \(22400 \mathrm{mL}\) volume is occupied by 1 mole of

\(O_{2}\) at \(STP\)

Thus, \(5600 \mathrm{mL} O_{2}\) means \(=\frac{5600}{22400} \mathrm{mol} O_{2}\)

\(=\frac{1}{4} m o l O_{2}\)

\(\therefore\) Weight of \(\mathrm{O}_{2}=\frac{1}{4} \times 32=8 \mathrm{g}\)

According to problem,

Equivalents of \(\mathrm{Ag}=\) Equivalents of \(\mathrm{O}_{2}\)

\(\begin{array}{l} {=\frac{\text {Weight of } A g}{\text {Equivalent wor}} \mathrm{Ag}} \\ {=\frac{\mathrm{W}_{O_{2}}}{\text { Equivalent weight of } O_{2}}} \end{array}\)

\(\frac{W_{A g}}{\frac{M_{g}}{1}}=\frac{W_{O_{2}}}{\frac{M_{0_{2}}}{4}}\)

\(\therefore \quad \frac{W_{A g}}{108} \times 1=\frac{8}{32} \times 4\)

\(\left[\because 2 H_{2} O \rightarrow O_{2}+4 H^{+}+4 e^{-}\right]\)

\(\Rightarrow \quad W_{A g}=108 g\)