Răspuns:
p= mpx100/m,impur
mp=95×227,3/100=215,935g Al
MAl=27g/mol
MFe2O3=2×56+3×16=160g/mol
2×27gAl..........160gFe2O3
2Al + Fe2O3 → Al2O3 + 2Fe
215,935gAl......xgF2O3
x=215,935×160/2×27=calculeaza
calculam ne de moli
MAl=27g/mol
-calculez numar de moli Al
n=m/M =215,935/27=7,99moli Al
2mol....... ........1mol
2Al + Fe2O3---> Al2O3 + 2Fe
7,99moli.......ymoliFe2O3
y=7,99×1/2=3.99 moli Fe2O3
Explicație: