M FeCO3= 56+12+3x16
=56+12+48
=116
1 mol FeCO3 = 116 g
M FePO4=56+31+4x16
=56+31+64
=151
1 mol FePO4 = 151 g
M FeNO3 =56+14+3x16
=56+14+48
=118
1 mol FeNO3 = 118 g
M FeOH=56+16+1
=73
1 mol FeOH =73 g
M FeCl =56+35,5
=91,5
1 mol FeCl =91,5 g
M FeS =56+32
=88
1 mol FeS =88 g
M FeO =56+16
=72
1 mol FeO =72 g
Sper ca-i inteles ce-am facut eu aici.
Baftaa !!