Răspuns:
Fe(CN)₂ = cianură de fier (2)
1. r.a. Fe:C:N = 1:2:2
2. r.m. Fe:C:N = A(Fe) : 2A(C) : 2A(N) = 56 : 2 × 12 : 2 × 14 = 56:24:28 = 14:6:7
3. μ(Fe(CN)₂) = A(Fe) + 2A(C) + 2A(N) = 56 + 2 × 12 + 2 × 14 = 108 g/mol
4.
108g Fe(CN)₂.....56g Fe.....24g C.....28g N
100g Fe(CN)₂.....x g Fe.....y g C.....z g N
x = 100 × 56 : 108 = 51,85% Fe
y = 100 × 24 : 108 = 22,22% C
z = 100 × 28 : 108 = 25,92% N
5.
1 mol Fe(CN)₂..........108g Fe(CN)₂
w mol Fe(CN)₂..........500g Fe(CN)₂
w = 500 : 108 = 4,62 mol Fe(CN)₂
6.
1 mol Fe(CN)₂..........108g Fe(CN)₂
3 mol Fe(CN)₂..........a g Fe(CN)₂
a = 3 × 108 = 324g Fe(CN)₂
-Mihai
