$\begin{cases}CuCl_2:0,05 mol \\ NaCl:0,25 m \end{cases}$
$CuCl_2\rightarrow Cu+Cl_2\uparrow$$2NaCl+2H_2O\rightarrow 2NaOH+H_2+Cl_2$
$2Al+2NaOH+2H_2O\rightarrow 2NaAlO_2+3H_2$
$\Rightarrow n_{Cu}=\frac{5.t_1}{2.96500}\Rightarrow t_1=1930 s\Rightarrow t_2=1750 s$
$\Rightarrow n_{NaOH}=\frac{5.1750}{1.96500}=0,09 mol$
$\Rightarrow m=2,43 gam$