Question

calculati masa moleculara,raportul atomic, raportul de masa,compozitia procentuala pentru Sulfat de aluminu si acid boric.
Urgent!!!!!!!!!

Answer 1

MAl₂(SO₄)₃ = 2×AAl + 3(AS+4×AO) = 2×27+3(32+4×16) = 342g/mol

raport atomic = 2:3:12 
raport masic = 54:96:192 = 9:16:32 

%Al = mAl/M×100 = 54/342×100 = 15,789
%S = 96/342×100 = 28,07
%O = 192/342×100 = 56,14

MH₃BO₃ = 3×AH + AB + 3×AO = 3×1+11+3×16 = 62g/mol

raport atomic = 3:1:3
raport masic = 3:11:48

%H = mH/M×100 = 3/62×100 = 4,838
%B = 11/62×100 = 17,74
%O = 48/62×100 = 77,419