3NaOH+FeCl3=Fe(OH)3+3NaCl
130 g sol FeCl3(ms)
c=5% c/100=md/ms=>md=c x ms/100=5x130/100=5x1,3=6,5g FeCl3
M FeCl3=56+35,5x3=56+106,5=162,5g/mol
numarul de moli=6,5/162,5=0,04 moli
1 mol FeCl3..................3 moli NaOH
0,04..................................0,12 moli NaOH
M NaOH=40g/mol
m=0,12x40=4,8g NaOH