A reactor is almost never charged in exact proportion. One reactant is used up while some of the other is still there, and at that moment the reaction stops. The reactant that is gone is the limiting reagent: it sets how much product the batch can make. The other is in excess, and some of it is still in the vessel when the drain valve opens.

The test takes two divisions and a look. Put each charge into moles, then divide each amount by its own coefficient from the balanced equation. The smaller result belongs to the limiting reagent. For $2\,\mathrm{NaOH} + \mathrm{H_2SO_4}$ with 900 mol of caustic soda and 500 mol of acid: $900 / 2 = 450$ and $500 / 1 = 500$, so the caustic soda runs out first, even though more moles of it were charged.

Then the arithmetic worth a formula. $n_{\text{excess}} = n_B - \dfrac{b}{a}\,n_A$, read aloud n-excess equals n-B minus b over a times n-A. The convention, stated once: A is the limiting reagent, B is the reagent in excess. $n_A$ is the amount of A charged and $n_B$ the amount of B charged, both in moles. $a$ and $b$ are their coefficients in the balanced equation, bare whole numbers. The term $\dfrac{b}{a}\,n_A$ is the amount of B the reaction uses, and $n_{\text{excess}}$ is the amount of B left unreacted, in moles. It is the quantity you are solving for.

The relation also checks your choice of A. Label the wrong reactant as limiting and the leftover comes out negative. That is not an error in the arithmetic. It is the calculation telling you the batch would need more B than was charged, so B is the one that runs out. Swap the labels and run it again.

Now the myth, because it costs real batches: the smaller mass runs out first. It does not. An equation counts moles and has never weighed anything. 49 kg of sulphuric acid is only 500 mol, while 44 kg of caustic soda is 1100 mol, because the caustic molecule is so much lighter. The acid is the bigger mass on the floor and it is still the reactant that runs out. Moles over coefficient decides it, every time.

Plants choose their excess on purpose. The cheap or harmless reactant is charged over the mark so that the expensive or troublesome one is completely used up, and the size of that margin is written into the procedure. Run forward, the relation tells you what will be left. Run backward, $n_B = n_{\text{excess}} + \dfrac{b}{a}\,n_A$, it tells you what to charge.