\(n_{FeCl_3}=\dfrac{m}{M}=\dfrac{16,25}{162,5}=0,1mol\)
FeCl3+3NaOH\(\rightarrow\)Fe(OH)3\(\downarrow\)+3NaCl
0,1........0,3............0,1...........0,3
\(m_{Fe\left(OH\right)_3}=0,1.107=10,7gam\)
\(m_{dd_{NaOH}}=\dfrac{0,3.40.100}{10}=120gam\)
\(m_{dd}=16,25+120-10,7=125,55gam\)
C%NaCl=\(\dfrac{0,3.58,5.100}{125,55}\approx\)14%