Răspunsuri

2016-11-13T17:32:32+02:00
8Al+3Fe3O4=4Al2O3=9Fe
V(niu)=m/M
M(Fe)=56 g/mol
Niu(Fe)=5,6g/56g/mol=0,1mol
Niu(Al)=(8mol*0,1mol)/9mol=0,09mol
M(Al)=27g/mol
m(Al)=27g/mol*0,09mol=2,43g
Niu(Fe3O4)=(3mol*0,1mol)/9mol=0,03mol
M(Fe3O4)=232g/mol
m(Fe3O4)=0,03mol*232g/mol=6,96g
Rs: m(Al)=2,43g
      m(Fe3O4)=6,96g
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