Question

When $FeC{l}_3$ is treated with $K_4\left[Fe\right(CN{)}_6]$, a blue coloured compound is formed. What are the total number of atoms per formula molecule? (supposing no K is present in blue coloured compound)

Answer 1

The corresponding reaction is:
$4FeC{l}_3+3K_4\left[Fe\right(CN{)}_6]\to F{e}_4[Fe(CN{)}_6{]}_3+12KCl$
The blue colour is due to $F{e}_4\left[Fe\right(CN{)}_6{]}_3$.
Therefore, total number of atoms per formula molecule are 43.