\(\begin{array}{l}
a)\\
NaOH+HCl\to NaCl+H_2O\\
n_{NaOH}=\frac{16}{40}=0,4(mol)\\
\to n_{NaCl}=n_{NaOH}=0,4(mol)\\
\to m_{NaCl}=0,4.58,5=23,4(g)\\
b)\\
m_{dd\,spu}=16+1,1=17,1(g)\\
C\%_{NaCl}=\frac{23,4}{17,1}.100\%>100\%
\end{array}\)
Xem lại \(m_{dd\,HCl}\)