Jaka byla masa probki Mn2O3, jezelj w wyniku redukcji otrzymano 0,312g Mn3O4 (MnO*Mn2O3)?
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M = 3*55g + 4*16g = 229g
229g(Mn₃O₄)----------165g (Mn)
0,312g---------------------x
X=0,312*165/229 =0,2248g
Mn₂O₃
M= 55g*2 + 16g*3 = 158g
158g(Mn₂O₃) ----110g(Mn)
x-------------------0,2248g
x= 0,2248*110/158 = 0,1565g