\(\begin{array}{l}
1)\\
\text{PTHH:}\ 2Na + 2{H_2}O \to 2NaOH + {H_2}\\
\text{Ta có:}\ nNa = \dfrac{{2,3}}{{23}} = 0,1\,mol \to n{H_2} = \dfrac{{0,1}}{2} = 0,05\,mol\\
\to V{H_2} = 0,05 \times 22,4 = 1,12l\\
2)\\
a)\\
\text{PTHH:}\ Zn + 2HCl \to ZnC{l_2} + {H_2}\\
nZn = \dfrac{{6,5}}{{65}} = 0,1\,mol \to n{H_2} = nZn = 0,1\,mol\\
\to V{H_2} = 0,1 \times 22,4 = 2,24l\\
b)\\
\text{Ta có:}\ nHCl = 2nZn = 0,2\,mol \to C_M(HCl) = \dfrac{{0,2}}{{0,1}} = 2M\\
3)\\
a)\\
\text{PTHH:}\ Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
nZn = \dfrac{{6,5}}{{65}} = 0,1\,mol\\
n{H_2} = nZn = 0,1\,mol \to V{H_2} = 0,1 \times 22,4 = 2,24l\\
b)\\
n{H_2}S{O_4} = Zn = 0,1\,mol \to C_M(H_2SO_4) = \dfrac{{0,1}}{{0,1}} = 1M\\
4)\\
nCuO = \dfrac{{48}}{{80}} = 0,6\,mol\\
n{H_2} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
\text{PTHH:}\ CuO + {H_2} \to Cu + {H_2}O\\
\text{Ta thấy:}\ \dfrac{{0,6}}{1} > \dfrac{{0,15}}{1} \Rightarrow nCu = n{H_2} = 0,15\,mol\\
\to mCu = 0,15 \times 64 = 9,6g
\end{array}\)