Question

Fe(CN)2
De aflat:
1.RA
2. RM
3 miu
4 compoziția
5 nr de moli
6 masa a

Answer 1

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