Дано: Решение:
m(NaOHр-р)=50г AlCl3 + 3NaOH(р-р) = Al(OH)3↓ + 3NaCl
w(NaOH)=10% n(AlCl3) = 3n(NaOH)
m(AlCl3)-? n(NaOH) = m/M = 5 / 23 + 17 = 5 / 40 = 0,125моль
m(NaOH) = 50 * 0.1 = 5г
n(AlCl3) = n(NaOH) / 3 = 0.042моль
m(AlCl3) = n * M = 0.042 * 133.5 = 5.6г