How can I know when I have added an excess of magnesium oxide? The following instructions were used to prepare magnesium sulfate crystals, $\ce{MgSO4 . 7H2O}$.


Measure $50~ \mathrm{cm^3}$ of dilute sulfuric acid into a beaker and warm the solution.  
Using a spatula, add some magnesium oxide and stir the mixture. Continue 
adding the magnesium oxide until excess is present.  
Separate the excess magnesium oxide from the solution of magnesium 
sulfate.  
Heat the solution until crystals form. Obtain the crystals and dry them.


The question I have an issue with is this: How would you know when excess magnesium oxide is present in step 2?

The answer key says that I would know when no more solid (that is, no more magnesium oxide dissolves). However, since we are talking about dilute sulfuric acid, there should be water, and I read on the internet that $\ce{MgO}$ is soluble in water.
$\ce{MgO}$ is insoluble in water (solubility: 0.0086 g/100 mL). On the other hand, $\ce{MgSO4}$ is soluble in water (solubility: 25.5 g/100 mL).

The reaction that is taking place is an acid-base reaction:

$$\ce{MgO + H2SO4 -&gt; MgSO4 + H2O}$$

Now for your question, how would you know when excess $\ce{MgO}$ is present? Here you are trying to make $\ce{H2SO4}$ the limiting reagent. When all the $\ce{H2SO4}$ has reacted, no more $\ce{MgO}$ can further react. Therefore, you will see solid in your reaction mixture that cannot be further dissolved.

At the end of your reaction,

$$\ce{MgO_{excess} + H2SO4 -&gt; MgSO4 + H2O + MgO_{remaining}}$$