\(PTHH:H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
Ban đầu 0,1________0,8
Phản ứng 0,1_________ 0,2 ______0,1
Dư ______ 0 ______ 0,6
\(m_{NaOH}=160.20\%=32g\)
\(n_{NaOH}=\frac{32}{23+17}=0,8\left(mol\right)\)
\(m_{H2SO4}=200.4,9\%=9,8\%\)
\(n_{H2SO4}=\frac{9,8}{32+2+16.4}=0,1\left(mol\right)\)
\(C\%_{Na2SO4}=\frac{0,1.\left(23.2+32+16.4\right)}{200+160}.100\%=3,94\%\)
\(C\%_{NaOH_{Du}}=\frac{0,6.\left(23+17\right)}{200}.100\%=65,16\%\)