What is formed when you leave iron(II) sulfate in plain air? What is formed when you leave iron(II) sulfate in plain air? 

Knowing that iron(II) is easily oxidised to iron(III), and assuming that the reactive component of air is oxygen, I solved it this way:

$$\ce{FeSO4 + O2 -&gt; Fe2O3 + SO2}$$

But in my textbook it's given that iron(III) sulfate, $\ce{Fe2(SO4)3}$, is formed. Is this because $\ce{Fe2O3 + SO2}$ react to form $\ce{Fe2(SO4)3}$?
Usually, the original iron(II) sulfate is present as green $\ce{FeSO4.7H2O}$.

Oxygen can oxidize $\ce{Fe(II)}$ salts to $\ce{Fe(III)}$ salts; for example, at $\mathrm{pH}=0$:

$$\begin{alignat}{2}
\ce{[Fe(H2O)6]^3+ + e- \;&amp;&lt;=&gt; [Fe(H2O)6]^2+}\quad &amp;&amp;E^\circ = +0.771\ \mathrm{V}\\
\ce{O2 + 4H+ + 4e- \;&amp;&lt;=&gt; 2H2O}\quad &amp;&amp;E^\circ = +1.229\ \mathrm{V}
\end{alignat}$$

$\ce{Fe(II)}$ is even easier oxidized under alkaline conditions; for example, at $\mathrm{pH}=14$:

$$\begin{alignat}{2}
\ce{FeO(OH) + H2O + e- \;&amp;&lt;=&gt; Fe(OH)2 + OH-}\quad &amp;&amp;E^\circ = -0.69\ \mathrm{V}\\
\ce{O2 + 2H2O + 4e- \;&amp;&lt;=&gt; 4OH-}\quad &amp;&amp;E^\circ = +0.401\ \mathrm{V}
\end{alignat}$$

However, you cannot simply oxidize iron(II) sulfate $\left(\ce{FeSO4}\right)$ to iron(III) sulfate $\left(\ce{Fe2(SO4)3}\right)$ in dry air since you would need additional sulfate to balance the equation:

$$\ce{2FeSO4 + SO4^2- -&gt; Fe2(SO4)3 + 2e-}$$

Nevertheless, $\ce{Fe(II)}$ can be oxidized to $\ce{Fe(III)}$.

The resulting $\ce{Fe(III)}$ is subject to hydrolysis. By way of comparison, the ion $\ce{[Fe(H2O)6]^3+}$ is only stable under strong acidic conditions. Already at $\mathrm{pH}=0{-}2$, it turns into yellow $\ce{[Fe(OH)(H2O)5]^2+}$ and begins to form $\ce{[Fe(OH)2(H2O)4]+}$:

$$\begin{align}
\ce{[Fe(H2O)6]^3+ \;&amp;&lt;=&gt; [Fe(OH)(H2O)5]^2+ + H+}\\
\ce{[Fe(OH)(H2O)5]^2+ \;&amp;&lt;=&gt; [Fe(OH)2(H2O)5]+ + H+}\\
\end{align}$$

Further addition of base causes precipitation of amorphous iron(III) hydroxide.

Accordingly, the likely product when iron(II) sulfate is oxidized is basic iron(III) sulfate, i.e. approximately:

$$\ce{4FeSO4 + O2 + 2H2O -&gt; 4Fe(OH)SO4}$$

However, the real weathering and aging of iron(II) sulfate in dry air actually yields a mixture of various compounds, including iron(III) sulfate and iron(III) oxide-hydroxide.

You may observe this reaction in some iron fertilizers for lawns. The fresh product typically contains green iron(II) sulfate, which gradually becomes yellow.