Question

what is the ratio of concentrated NH4SCN and [Fe (H2O]^3+ is taken for better yield of [Fe (SCN)6]^-3​

Answer 1

Answer:

The ratio of concentrated $NH_{4}SCN$ and $[Fe(H_{2}O)_{6}]^{3+}$ is 6 : 1 for better yield of $[Fe(SCN)_{6}]^{3-}$

Explanation:

The reaction of $NH_{4}SCN$ and $[Fe(H_{2}O)_{6}]^{3+}$ is

6$NH_{4}SCN$ + $[Fe(H_{2}O)_{6}]^{3+}$⇄ $[Fe(SCN)_{6}]^{3-}$ + 6$NH^{+}_{4}$ + 6$H_{2}O$

6 mole of $NH_{4}SCN$ react with 1 mole of $[Fe(H_{2}O)_{6}]^{3+}$ to prepare 1 mole  $[Fe(SCN)_{6}]^{3-}$.

So, The ratio of concentrated $NH_{4}SCN$ and $[Fe(H_{2}O)_{6}]^{3+}$ is 6:1 for better yield of $[Fe(SCN)_{6}]^{3-}$.