Question

- calculeaza masa hidroxidului de aluminiu Al(OH)3 format la interactiunea sulfatuluide aluminiu cu 7,4 g de var stins Ca(OH)2

Answer 1

MAl2(SO4)3=27*2+3(32+16*4)=342g/mol
MCa(OH)2=40+2(16+1)=74g/mol
MAl(OH)3=27+3(16+1)=78g/mol
342g.........3*74=222g.........2*78=156g
Al2(SO4)3 + 3Ca(OH)2 --> 2Al(OH)3 + 3CaSO4
x=11.4g.....7.4g....................y=156*(11.4+7.4)/222+342=5.2g Al(OH)3