\(n_{HCl}=\dfrac{300.7,3\%}{36,5}=0,6\left(mol\right)\)
\(n_{NaOH}=\dfrac{200.4\%}{40}=0,2\left(mol\right)\)
\(NaOH+HCl\rightarrow NaCl_2+H_2O\)
0,2 0,2 0,2 0,2
Ta thấy \(\dfrac{0.6}{1}>\dfrac{0.2}{1}\Rightarrow HCl.dư,NaOH.đủ\)
\(m_{HCl\left(dư\right)}=\left(0,6-0,2\right).36,5=14,6\left(g\right)\)
\(m_{H_2O}=0,2.18=3,6\left(g\right)\)
\(m_{dd_{NaCl}}=\left(300+200\right)-3,6=496,4\left(g\right)\)
\(C\%_{HCl}=\dfrac{14.6}{200+300}.100=2,92\%\)
\(C\%_{NaCl}=\dfrac{0,2.58,5}{200+300}.100=2,34\%\)