Can the limiting reactant be the one with the highest starting mass?

Yes! If a reactant is very heavy (high molar mass), a large mass in grams might actually be a very small number of moles. It is very common for the heaviest starting material to be the limiting reactant.

What happens to the excess reactant?

The excess reactant is simply left over in the reaction vessel. Once the limiting reactant is completely consumed, the reaction stops, leaving the unreacted portion of the excess reactant behind.

Do I always have to use the coefficients?

Yes. The coefficients tell you the required ratio of ingredients. If you don't divide by the coefficient (or use mole ratios to check product amounts), you are ignoring the recipe for the reaction.

How do you find the limiting reactant?

Finding the limiting reactant means figuring out which ingredient in a chemical reaction will run out first. Think of building bicycles: if you have 10 frames and 14 wheels, you can only build 7 bikes. The wheels run out first, so they limit your production, even though you still have frames left over. In chemistry, molecules don't combine gram-by-gram; they combine piece-by-piece. To find the limiting reactant, we must convert our starting materials into moles, compare them to the recipe (the balanced chemical equation), and see which one produces the smallest amount of product. The reactant that runs out first dictates exactly how much product you can make.

What is a limiting reactant?

By definition, the limiting reactant is the substance that is totally consumed when a chemical reaction is complete. The amount of product formed is entirely dependent on this reactant. Any other reactants that are not fully consumed are called excess reactants. Because elements and compounds have different molar masses, you cannot simply look at the starting mass in grams to know which one will run out first.

The step-by-step method

To find the limiting reactant, you generally follow three steps. First, ensure your chemical equation is balanced. Second, convert the given amounts of all reactants from grams (or liters) into moles. Third, divide the moles of each reactant by its stoichiometric coefficient from the balanced equation. The reactant with the smallest resulting number is your limiting reactant. This normalized number tells you how many 'reaction cycles' each reactant can support.

Where students slip up

The most common mistake is skipping the mole conversion and simply comparing the starting masses in grams. A heavier molecule might have a large mass but represent very few actual particles. Always convert to moles first. Another common trap is forgetting to divide by the coefficients. If a reaction needs 3 moles of hydrogen for every 1 mole of nitrogen, you must account for that 3-to-1 ratio before deciding which runs out first.

Worked through

Consider the synthesis of ammonia: $N_2 + 3H_2 \rightarrow 2NH_3$ If you start with 14.0 g of $N_2$ and 6.0 g of $H_2$ , what is the limiting reactant?

Step 1: Convert both starting masses to moles. Moles of $N_2$ = 14.0 g / 28.0 g/mol = 0.50 mol $N_2$ Moles of $H_2$ = 6.0 g / 2.0 g/mol = 3.0 mol $H_2$ Step 2: Divide each mole value by its coefficient from the balanced equation. For $N_2$ : 0.50 / 1 = 0.50 For $H_2$ : 3.0 / 3 = 1.0 Step 3: Compare the values. Since 0.50 is less than 1.0, Nitrogen ( $N_2$ ) will run out first. Therefore, $N_2$ is the limiting reactant.

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Where this comes from: OpenStax Chemistry 2e: Chapter 4 · Khan Academy: Stoichiometry Unit · Chemistry: The Central Science by Brown, LeMay, Bursten