Solution:
In \(15 L\) of \(H_{2}\) gas at \(S T P\), the number of molecules
$$
=\frac{6.023 \times 10^{23}}{22.4} \times 15=4.033 \times 10^{23}
$$
In \(5 L\) of \(N_{2}\) gas at STP, the number of molecules
$$
=\frac{6.023 \times 10^{22} \times 5}{22.4}=1.344 \times 10^{23}
$$
In \(0.5 \mathrm{~g}\) of \(\mathrm{H}_{2}\) gas, the number of molecules
$$
=\frac{6.023 \times 10^{23} \times 0.5}{2}=1.505 \times 10^{23}
$$
In \(10 \mathrm{~g}\) of \(O_{2}\) gas, the number of molecules
$$
=\frac{6.023 \times 10^{23} \times 10}{32}=1.882 \times 10^{23}
$$
Thus, \(15 \mathrm{~L}\) of \(\mathrm{H}_{2}\) gas at STP has the maximum number of molecules
