\(n_{NaOH}=\frac{40.20}{40}=0,2\left(mol\right)\)
\(n_{H2SO4}=\frac{78,4.25}{98}=0,2\left(mol\right)\)
PTHH:\(\text{a. 2NaOH+H2SO4-->Na2SO4+2H2O}\)
trước........0.2..............0.2
phản ung 0.2............. 0.1
dư...........0.............. 0.1...................0.1
\(\Rightarrow m_{Na2SO4}=0,1.142=14,2g\)
\(\text{b. mdd sau phản ứng=40+78.4=118.4}\)
\(C\%_{NA2SO4}=\frac{14,2}{118.4}=11,99\%\)
\(C\%_{H2SO5}=\frac{0,1.98}{118.4}=8,27\%\)