Răspuns:
151,25g Fe(NO₃)₃
Explicație:
oxid de fer (3) - Fe₂O₃
acid azotic - HNO₃
M(Fe₂O₃) = 2 × A(Fe) + 3 × A(O) = 2 × 56 + 3 × 16 = 160 g/mol
M(Fe(NO₃)₃) = A(Fe) + 3 × A(N) + 9 × A(O) = 56 + 3 × 14 + 9 × 16 = 242 g/mol
50g x
Fe₂O₃ + 6HNO₃ → 2Fe(NO₃)₃ + 3H₂O
160g 2 × 242g
x = 50 × 2 × 242 : 160 = 151,25g Fe(NO₃)₃
-Mihai