ta có mNaOH= 400. 30%= 120( g)
\(\rightarrow\) nNaOH= 120/ 40= 3(mol)
PTPU
NaOH+ HCl\(\rightarrow\) NaCl+ H2O
3.............3..........3............
a) ta có mNaCl= 3. 58,5= 175,5( g)
mdd NaCl= 400+ 100= 500( g)
\(\Rightarrow\) C%NaCl= \(\dfrac{175,5}{500}\). 100%=35,1%