Ờ, có bạn nhắc, nhầm nhé :)
\(C\%_{NaOH}=\frac{0,8.40}{500}.100\%=6,4\left(\%\right)\)
\(n_{HCl}=\frac{7,3.200}{100.36,5}=0,4\left(mol\right)\\ n_{H_2SO_4}=\frac{9,8.200}{100.98}=0,2\left(mol\right)\\ PTHH:NaOH+HCl\rightarrow NaCl+H_2O\\ PTHH:2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ \sum n_{NaOH}=0,4+0,2.2=0,8\left(mol\right)\\ C\%_{NaOH}=\frac{0,8.40}{200+500}.100\%=4,57\left(\%\right)\\ C\%_{ddspu}=\frac{0,4.58,5+0,2.142}{200+500}.100\%=7,4\left(\%\right)\)