inorganic chemistry - Are these reaction equations for the formation of the brown ring complex correct?

\begin{align} \ce{NO3- + 3Fe^2+ + 4H+ &-> NO + 3Fe^3+ + 2H2O}\tag{1}\\ \ce{[Fe(H2O)6]^2+ + NO &-> \underset{\text{(brown)}}{[Fe(H2O)5(NO)]^2+} + H2O}\tag{2} \end{align}

The above are the reactions given in my book [1] for the qualitative analysis of nitrate ion (formation of the brown ring complex). I think that the reactions above are incorrect. The first one shows formation of $\ce{Fe^3+}$, which should be attacked by excess of $\ce{H2O}$ to give ferrum(III) hexaaquasulphate, not ferrum(II) hexaaquasulphate. Now that should be attacked by $\ce{NO}$ as a ligand and give $\ce{[Fe(H2O)6NO]SO4}$, where the oxidation number of $\ce{Fe}$ is being $3+$.

Here the reactions show that the $\ce{Fe}$ already present gives $\ce{[Fe(H2O)6]^2+}$ and then $\ce{NO}$ acts as a neutral ligand to form the complex. But by the reactions shown in the book, even after formation of the brown ring complex $\ce{Fe^3+}$ is still present in the solution, which according to me shouldn't be because if $\ce{Fe^3+}$ still remains in the solution then water should also have attacked it as a ligand to give $\ce{[Fe(H2O)6NO]SO4}$.

I need some clarity on this. Are the reactions given in the book correct?

Edit: After reading the answers, I do get that the hexaaqua iron(II) will be formed and the $\ce{Fe^3+}$ will be attacked by water to give $\ce{[Fe(H_2O)_6]^3+}$ but why won't $\ce{NO}$ take electrons from the oxidation of iron ($\ce{Fe^2+ -> Fe^3+ + e- }$), make $\ce{NO-}$ and attack the above and substitute one water molecule and give more of the brown ring complex?

1. NCERT. Chemistry: Textbook for class XII; National Council of Educational Research and Training: New Delhi, 2007. ISBN 978-81-7450-648-1.