Zadanie 1
= 174 g/mol
= 32 g/mol
174 g ----- 100 %
32 g ----- x
x = 18,39 %
Zadanie 2
Fe(OH)₂ --T--> FeO + H₂O
masa substratów = masa produktów
89,8 g
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Zadanie 1
= 174 g/mol
= 32 g/mol
174 g ----- 100 %
32 g ----- x
x = 18,39 %
Zadanie 2
Fe(OH)₂ --T--> FeO + H₂O
masa substratów = masa produktów
89,8 g