Excess Reagent Remaining

Once you know which reactant ran out, the interesting question is what is left of the other one. Subtract what the limiting reagent consumed from what you charged: n_excess = n_B − (b/a)n_A. It is the second half of a limiting-reagent problem and the half that gets rushed, even though in industry it is the half that matters — leftover reagent is cost, and it is usually also a separation problem.

Worked: N₂ + 3H₂ → 2NH₃, charged with 5.00 mol of nitrogen and 20.0 mol of hydrogen. Nitrogen runs out first, so it is the limiter. Hydrogen consumed = (3/1) × 5.00 = 15.0 mol, leaving 20.0 − 15.0 = 5.00 mol of hydrogen sitting unreacted. In an ammonia plant that leftover is not thrown away — it is separated from the ammonia and recycled, which is why the real recycle loop runs a large excess of both gases past the catalyst.

A negative answer here is not an error; it is a diagnosis. It means B ran out before A did, so you labelled the wrong reactant as limiting. Swap the two, run it again, and the question you actually wanted answered is how much of A is left over. The solver says so rather than blocking the calculation, because the negative number carries the information.

The limiting-reagent test itself is deliberately not a page on this site: it is a comparison rather than a relation. Divide each reactant's moles by its coefficient — n_A/a against n_B/b — and the smaller quotient limits. Two divisions and a look. There is no inverse to it, because the forward direction throws away the losing reactant's number entirely, and a calculator that pretends otherwise would be lying about what it knows.

To turn the leftover moles back into something you can weigh, multiply by molar mass. Chemists usually run the cheap or easily removed reactant in excess on purpose, to push a reversible reaction forward — Le Chatelier's principle applied with a purchase order.