Odpowiedź :
2 HCOOH + Mg = (HCOO)₂Mg + H₂
xg kwasu --------------- 11,4g soli
92g kwasu ---------------- 114 g soli
x = 92*11,4/114 = 9,2 g
xg kwasu --------------- 11,4g soli
92g kwasu ---------------- 114 g soli
x = 92*11,4/114 = 9,2 g
2*46g...............114g
2HCOOH+ Mg---> (HCOO)2Mg+ H2
......x.................11,4g
x= 2*46g*11,4g/114g= 9,2g
Użyto 9,2grama kwasu metanowego