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n H2O в р-ре AlCl3 = (20*(1-0.05))/18 = 1.0556 моль;
n AlCl3 = (20*0.05)/133.5 = 0.0075 моль;
1) Na + H2O = NaOH + 0.5H2 - вода в изб., обр-я 0.1304 моль NaOH;
2а) 3NaOH + AlCl3 = Al(OH)3 + 3NaCl;
2б) 4NaOH + AlCl3 = Na[Al(OH)4] + 3NaCl (не зная конц-и обр-я р-ра NaOH, предп-м, что обр-я именно такой комплекс);
мольное соотношение NaOH и AlCl3 равно 0.1304/0.0075 = 17.387 к 1, т.е. гарантированно будет обр-я компл-я соль по р-ии 2б);
т.о. в обр-я р-ре будут сод-я Na[Al(OH)4] массой 118*0.0075 = 0.885 г и NaOH массой 40*(0.1304-0.0075*4) = 4.016 г.