Răspuns:
K2CO3= carbonat de potasiu
M.K2CO3=
=2×AK+AC+3×AO= 2x39+12+3×16
=78+12+48
=138
Compoziția procentuala =
138gK2CO3....... 78gK.....12gC.....48gO
100gK2CO3.........xgK.......ygC.......zgO
138/100=78/x=12/y=48/z
X=78×100/138=56,521
Y=12×100/138=8,695
Z=48×100/138=34,782
