2 Al + 3 FeO ---> 3 Fe + Al2O3
a)2*27 g Al..........3 * 72 g FeO
x g Al..............160 g FeO x = 40 g Al
2*27 g Al.........102 g Al2O3
40 g Al.............y g Al2O3 (vei avea nevoie la subpunctul b) y = 75,555 g Al2O3
b) Al2O3 + 3 H2SO4 ---> Al2(SO4)3 + 3 H2O
102 g Al2O3.......3*98 g H2SO4
75,555 g Al2O3.........z g H2SO4 z = 217,777 g H2SO4 = md
c%=md/ms*100=>ms=217,777/46 * 100 =>ms = 473,43 g solutie