дано
m(AlCL3) = 25 g
------------------------
m(NaOH)-?
AlCL3+3NaOH-->Al(OH)3+3NaCL
M(AlCL3) = 133.5 g/mol
n(AlCL3) = m/M = 25 / 133.5 = 0.19 mol
n(AlCl3) = 3n(NaOH)
n(NaOH) = 3*0.19 = 0.57 mol
M(NaOH) = 40 g/mol
m(NaOH) = n*M= 0.57 * 40 = 22.8 g
ответ 22.8 г