b
\(2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\)

\(\mathrm{H}_{2}=10 \mathrm{gm}\)

\(\mathrm{O}_{2}=64 \mathrm{gm}\)

No. of moles of \(\mathrm{H}_{2}=\frac{10}{3}\)\(=5\) moles

\(\text { No. of moles of } \mathrm{O}_{2}=\frac{64}{32}\)\(=2 \mathrm{moles}\)

As \(1 \;mole\) of \(\mathrm{O}_{2}\) gives 2 moles of \(\mathrm{H}_{2} \mathrm{O}\)

\(\therefore 2 \;\mathrm{moles}\) of \(\mathrm{O}_{2}\) will give 4 mole of \(\mathrm{H}_{2} \mathrm{O}\)