1.procentul de O este 88,89.
transform raportul de mase in raport molar
n= m/A--> n,H:n,O= 11,11/1 : 88,89/16=11,11 /5,55
impart la cel mai mic = 2 : 1--- > H2O
2.mS: mO=1:1
nS:nO= 1/32 : 1/16, impart la cel mai mic,
= 1 : 2-------> SO2
M,SO2= (32gS+32gO)mol=63g/mol
63gSO2..........32gS.........32gO
100g.................X................Y.....................calculeaza !!!!!
SAU
(1+1)gSO2.........1gS........1gO
100g...................X...............Y.....................calculeaza !!!!
urmareste acest rationament si aplica-l la celalalt oxid al sulfului