Дано
m(ppa AlCL3) = 150 %
w(AlCL3) = 25%
--------------------------
m(LiOH)-?
m(AlCL3)= 150*25% / 100% = 37.5 g
AlCL3+3LiOH--->Al(OH)3+3LiCL
M(AlCL3) = 133.5 g/mol
n(AlCL3) = m/M = 37.5 / 133.5 = 0.28 mol
n(ALCL3) = n(Al(OH)3) =0.28 mol
M(Al(OH)3) = 78 g/mol
m(Al(OH)3) = n*M = 0.28*78 = 21.84 g
ответ 21.84 г