Question

What is a rate of concentrated NH4SCN and (Fe(H2O)6)3+is taken for better yield of (Fe(SCN)6)3-

Answer 1

Answer:

No one %%*692*

Explanation:

Answer 2

Given:  

• Concentrated   $NH_4SCN$ and $(Fe(H_2O)_6)^{3+}$ is given.

To Find:

• The ratio of concentrated   $NH_4SCN$ and $(Fe(H_2O)_6)^{3+}$ for better yield of $(Fe(SCN)6)^{3-}$

Solution:

The equation of $NH_4SCN$ and $(Fe(H_2O)_6)^{3+}$ is as follows

6 $NH_4SCN$+ $(Fe(H_2O)_6)^{3+}$ ⇄  $(Fe(SCN)6)^{3-}$+ $6NH_4^{+}$+ 6  $6H_2O$

From the above equation we can see that when 6 mole of $NH_4SCN$  react with 1 mole of $(Fe(H_2O)_6)^{3+}$  react then 1 mole of  $(Fe(SCN)6)^{3-}$ is formed.

Thus, The ratio of concentrated  $NH_4SCN$ and  $(Fe(H_2O)_6)^{3+}$is 6:1 for better yield of  $(Fe(SCN)6)^{3-}$