$n_{HCl}=0,3.0,1=0,03(mol)$
$n_{NaOH}=0,1.0,2=0,02(mol)$
$NaOH+HCl\to NaCl+H_2O$
$\dfrac{0,03}{1}>\dfrac{0,02}{1}$
$\to NaOH$ hết, $HCl$ dư
$V_{\rm dd spu}=300+100=400ml=0,4l$
$n_{NaCl}=n_{NaOH}=0,02(mol)$
$\to C_{M_{NaCl}}=\dfrac{0,02}{0,4}=0,05M$
$n_{HCl\rm dư}=0,03-0,02=0,01(mol)$
$\to C_{M_{HCl}}=\dfrac{0,01}{0,4}=0,025M$