$64.$

$Na+2HCl \to NaCl+H_2↑$

$Na_{dư}+H_2O \to NaOH +\dfrac{1}{2}H_2$

$n_{H_2}=0.15(mol)$

Bảo toàn $e:$

$n_{Na}=2n_{H_2}=0.3(mol)$

`=>`$m=0.3 \times 23=6.9(gam)$

$n_{Na}=0.3(mol)$

$n_{HCl}=\dfrac{100 \times 7.3\%}{100\% \times 36.5}=0.2(mol)$

`=>`$n_{NaOH}=n_{Na,dư}=0.3-0.2=0.1(mol);n_{NaCl}=0.2(mol)$

Bảo toàn khối lượng: $m_{dd}=100+6.9-0.15 \times 2=106.6(gam)$

$C\%(NaOH)=\dfrac{0.1 \times 40}{106.6} \times 100\%=3.75\%$

$C\%(NaCl)=\dfrac{0.2 \times 58.5}{106.6} \times 100\%=10.98\%