\(n_{NaOH}=\dfrac{250.9,6\%}{40.100\%}=0,6\left(mol\right)\)
\(NaOH+HCl--->NaCl+H_2O\)
0,6............0,6.....................0,6...........0,6
\(m_{HCl}=0,6.36,5=21,9\left(g\right)\)
\(C\%ddHCl=\dfrac{21,9}{150}.100\%=14,6\%\)
\(m_{NaCl}=0,6.58,5=35,1\left(g\right)\)
\(C\%=\dfrac{35,1}{150+250}.100\%=8,775\%\)