$1)$$NaOH+HCl\rightarrow NaCl+H_2O$
$\Rightarrow n_{NaCl}=\frac{14,04(20+30)}{58,5.100}=0,12 mol$
$\Rightarrow n_{NaOH}=0,12 mol\Rightarrow m_{NaOH}=4,8 gam\Rightarrow x=24$
$\Rightarrow n_{HCl}=0,12 mol\Rightarrow m_{HCl}=4,38 gam\Rightarrow y=14,6$