1) 2Al + 2NaOH + 6H2O = 2Na[Al(OH)4] + 3H2
2) Al2O3 + 2NaOH + 3H2O = 2Na[Al(OH)4]
n (H2) в 1) = 3.36/22.4 = 0.15 моль
m (Al) в 1) = 0.15 * 27 = 4.05 гр.
n (Al) = 4.05/27 = 0.15моль
зн. m (Al2O3) = 7.8 - 4.05 = 3.75 гр.
n (Al2O3) = 3.75/102 = 0.037моль
w (Al) = 4.05/7.8 * 100% = 51.9%
w (Al2O3) = 100% - 51.9% = 48.1%
n общ (NaOH) = n (Al) + 0.5n (Al2O3) =0.1685моль
m (NaOH) = 0.1685 * 40 = 6.74 гр.
m р-ра (NaOH) = 6.74/0.4 = 16.85 гр.
V (NaOH) = 16.85/1.4 = 12.04