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1 mol 2 mol
Mg + 2HCOOH → (HCOO)2Mg + H2
x 0.2 mol
x = 0.1 mol. Nadmiar magnezu (избыток магния).
M (HCOO)2Mg = 114 g/mol
2 mol 1 mol
Mg + 2HCOOH → (HCOO)2Mg + H2
0.2 mol x
x = 0.1 mol
114 g ---- 1 mol
x ---- 0.1 mol
x = 11.4 g
proszę :)