\(n_{HCl}=0,02.1,8=0,036\left(mol\right)\)
\(m_{dd.HCl}=20.1,8=36\left(g\right)\Rightarrow m_{H_2O}=36-0,036.36,5=34,686\left(g\right)\)
=> \(n_{H_2O}=\dfrac{34,686}{18}=1,927\left(mol\right)\)
PTHH: 2Na + 2HCl --> 2NaCl + H2
0,036---------->0,018
2Na + 2H2O --> 2NaOH + H2
1,927---------->0,9635
=> \(V_{H_2}=22,4.\left(0,018+0,9635\right)=21,9856\left(l\right)\)