nHCl=0,3*1=0,3 mol
nNaOH=\(\dfrac{4}{40}\)=0,1mol
PTHH: NaOH+HCl\(\rightarrow\)NaCl+H2O
Ta có \(\dfrac{0,3}{1}\)>\(\dfrac{0,2}{1}\)
Nên HCl dư
\(\Rightarrow\)nHCl dư=0,3-0,2=0,1mol
\(\Rightarrow\)mHCldư=0,1\(\times\)36,5=3,65g
Ta có
nNaCl=nNaOH=0,1mol
\(\Rightarrow\)mNaCl=0,1\(\times\)58,5=5,58