[tex]m_{NaOH}=?\\m_{sol. FeCl_{3}}=65g, c=\frac{25}{100}\\\\\\c=\frac{m_{d}*100}{m_{sol}}=>m_{d}=\frac{25*m_{sol}}{100}=\frac{25*65}{10000}=\frac{1625}{100}=16,25g (FeCl_{3})\\\\\\ x...............16,25g.............................y\\3NaOH + FeCl_{3}->3NaCl+Fe(OH)_{3}\\3*58,5g.....162g.........................107g\\\\x=\frac{3*58,5*16,25}{162}=17,6 g(NaOH)\\\\[/tex]
[tex]y=\frac{16,25*107}{162}=10,73g(Fe(OH)_{3})[/tex]
