The limiting reagent · printable packet

Which reagent runs out first, why it is not the one there is least of by mass, and what it allows.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to decide which of two weighed-out reagents runs out first — by turning each mass into moles and dividing by that reagent's coefficient, never by comparing the masses — and to use the limiting reagent alone to work out the amount and the mass of product a mixture can make. You will be able to say why the reagent in excess plays no part in the yield, and why an excess is often a deliberate choice.

2. What you already have

You can take one substance through a balanced equation to another. Every one of those calculations quietly assumed there was enough of everything else. This lesson removes the assumption, which is what happens the moment a real mixture is weighed out.

3. Words for this lesson

4. Divide the moles by the coefficient

Not the one there is less of by mass — that compares two different substances by a number that says nothing about how many particles there are. And not the one with the smaller number of moles either, because the equation may want three of one for every one of the other.

Term	What it means
Limiting reagent	The reactant that runs out first and so decides how much product forms.
In excess	Present in more than the reaction can use; some is left over.
Theoretical yield	The mass of product the limiting reagent allows.
Reagent	A substance added to make a reaction happen; here, a reactant.
Stoichiometric proportions	Reactant amounts in exactly the ratio of their coefficients.

That fraction answers a question worth saying out loud: how many times over could this reagent run the reaction as written, if everything else were unlimited? Whichever answer is smaller is the reagent that runs out, and the reaction runs that many times.

Then the rest is the last lesson. Take the smaller figure, multiply by the product's coefficient for the amount of product, and by the product's molar mass for its mass. The excess reagent plays no further part at all.

Mass never appears in the comparison, and that is the whole difficulty: masses are the numbers you were given, and they are the one thing that cannot be compared directly.

Another way: picture

A sandwich takes two slices of bread and one slice of cheese. With 10 slices of bread and 8 of cheese: bread makes $10/2 = 5$ sandwiches, cheese makes $8/1 = 8$. Five sandwiches, and three slices of cheese left over. Nobody weighs the bread.

Another way: steps

Turn each reagent's mass into moles, using its own molar mass.
Divide each amount by that reagent's coefficient in the balanced equation.
The smaller answer is the limiting reagent, and its value is how many times the reaction runs.
Multiply that by the product's coefficient for the moles of product.
Multiply by the product's molar mass for the mass of product.

5. The method, step by step, and how to check it

Balance the equation. The coefficients are half of the test, so they have to be right.

Convert each mass to moles. Each reagent through its own molar mass. Never compare the masses at this point or any other.

Divide each amount by its coefficient. This is the number of times that reagent alone could run the reaction.

Pick the smaller. That reagent limits. If the two figures are equal, the mixture is in exact proportion and both run out together.

Find the product from the limiting reagent only. Its figure times the product's coefficient gives moles of product; times the product's molar mass gives grams.

Find the leftover, if asked. The excess reagent used is the limiting figure times the excess reagent's coefficient; subtract that from what was supplied.

Check the work. Work out the product from the excess reagent too, just as a test: it should come out larger, because the excess could have made more if it had had a partner. If it comes out smaller, you picked the wrong reagent. And check mass: the reactants consumed should weigh the same as the products made.

6. Why each step is allowed

Converting to moles is required because the equation's coefficients count particles, and only moles are a count.

Dividing by the coefficient is the right comparison because it puts both reagents on the same footing: how many complete runs of the equation each could support. A reagent with a coefficient of 3 needs three of its particles for every run, so its supply is used up three times as fast.

Taking the smaller figure is required because a reaction cannot run a fraction more times than its scarcest ingredient allows. Once that reagent is gone, the reaction stops, however much of the other remains.

Ignoring the excess for the yield is allowed because the excess was never the constraint. Adding more of something already in surplus changes nothing.

The excess-reagent check works because the excess, by definition, could support more runs than actually happened. A product calculated from it is the yield of a reaction that never got to run.

7. Why not just compare the moles

Take $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$ with 1 mol of nitrogen and 2 mol of hydrogen. There is twice as much hydrogen, and hydrogen still runs out: the equation wants three hydrogens per nitrogen, and only two are there.

The hydrogen's figure is smaller, so the reaction runs 0.67 times over and makes $0.67 \times 2 = 1.33$ mol of ammonia. A third of a mole of nitrogen is left unreacted.

Comparing the moles alone would have said nitrogen. Comparing the masses — 28 g against 4 g — would have said hydrogen, and been right by accident, which is worse.

8. How much is left over

Once you know how many times the reaction ran, the leftover follows from the excess reagent's own coefficient. Each run uses that many moles of it, so the amount used is the limiting figure times the excess reagent's coefficient, and the leftover is what was supplied minus what was used.

Take $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$ with 1 mol of nitrogen and 2 mol of hydrogen again. Hydrogen limits, and the reaction runs $2/3$ of a time. Nitrogen used: $2/3 \times 1 = 0.67$ mol. Nitrogen left: $1 - 0.67 = 0.33$ mol, which is $0.33 \times 28 \approx 9.3$ g.

The leftover is a useful check on the whole calculation. Add the mass of product to the mass of leftover excess, and the total must equal the mass of everything you started with. Here the ammonia is $1.33 \times 17 \approx 22.7$ g, the leftover nitrogen 9.3 g, and together they make 32 g — the 28 g of nitrogen plus 4 g of hydrogen that went in. If the totals disagree, something earlier was wrong.

9. What the excess is for

Because reactions do not go to completion on their own. An excess of the cheaper reagent pushes more of the expensive one to react, and it is a standard, deliberate choice in industry: use a surplus of whatever is cheap, easy to recover or harmless, so that none of the costly reagent is wasted.

A car engine is a familiar case. Engines are tuned to run with a little more air than the fuel strictly needs, so that every drop of gasoline burns completely to carbon dioxide rather than to carbon monoxide. Air is free; gasoline is not, and carbon monoxide is dangerous.

	amount	coefficient	$n/c$
$\mathrm{N_2}$	1 mol	1	1
$\mathrm{H_2}$	2 mol	3	0.67

So an excess is usually a decision rather than a mistake, and the question which reagent did they choose to run in excess, and why often tells you more about a process than the yield does.

10. In the world: welding rail on the American railroad network

Most of the roughly 140,000 miles of freight railroad in the United States is continuous welded rail, and many of the joints are made in the field with thermite. A crew clamps a ceramic mold around the gap between two rails, fills a crucible above it with a powdered mixture of iron oxide and aluminum, and ignites it. The reaction $\mathrm{Fe_2O_3 + 2Al \rightarrow Al_2O_3 + 2Fe}$ reaches over 4,000 °F, and the molten iron pours down into the gap and fuses the rails into one.

The charge is a limiting-reagent problem designed in advance. The manufacturer wants every gram of aluminum to react, because unburned aluminum left in the weld weakens it. So the aluminum is made the limiting reagent and the iron oxide is supplied in a slight excess. A typical portion of about 10 kg might hold 7.5 kg of iron oxide and 2.5 kg of aluminum. Moles over coefficients: $7500 \div 160 \approx 47$ for the oxide and $2500 \div 27 \div 2 \approx 46$ for the aluminum — nearly matched, with the aluminum just short.

The iron the weld receives is set by the aluminum: about $46 \times 2 \times 56 \approx 5.2$ kg. Comparing masses would have suggested the opposite — the oxide is three times heavier than the aluminum — and a crew trusting that would expect the wrong reagent to run out. The aluminum oxide floats to the top as slag and is chipped away, and the rail is ground smooth. The whole job takes under an hour, and the equation decided the recipe before the crew arrived.

11. In the world: why a gas stove has an air shutter

A burner mixes natural gas with air before it lights. With too little air, oxygen becomes the limiting reagent and the flame turns yellow and sooty; the shutter lets in enough that methane always limits, and the flame burns blue.

12. Where this goes wrong

Reading less mass as less reagent. The limiting reagent is not the one there is least of by mass. It is the one whose moles, divided by its coefficient, come out smallest — and the balanced equation is what supplies the coefficient, which is why balancing comes first.

Comparing masses. The error this lesson exists for. Two masses of different substances cannot be compared: they are counts of different things at different weights. Convert to moles first, always.

Comparing moles without the coefficients. Closer, and still wrong whenever the coefficients differ. Two moles of hydrogen is not enough for one mole of nitrogen.

Multiplying by the coefficient instead of dividing. Ask what the fraction means: how many times can this reagent run a reaction that needs $c$ of it? That is $n$ divided by $c$, never $n$ times $c$.

Using the excess reagent for the yield. Once the limiting reagent is identified, the other one is finished with. Calculating the product from the excess gives a yield larger than the reaction can possibly produce.

13. Where the heavier reagent runs out

14. Where the lighter reagent runs out

15. Thermite, with the leftover

16. Your turn: $\mathrm{2Na + Cl_2 \rightarrow 2NaCl}$, with 46 g of sodium and 142 g of chlorine.

17. Guided practice

In $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$, a mixture contains $28$ g of nitrogen and $12$ g of hydrogen. Which reagent runs out first?

18. Guided practice

Complete the worked solution: in $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$, twenty-eight grams of nitrogen meet nine grams of hydrogen. Nitrogen gas is twenty-eight grams per mole, hydrogen gas two, ammonia seventeen. The nitrogen's moles over coefficient is one. Find the hydrogen's moles, its moles over coefficient, and the mass of ammonia made.

19. Guided practice

For the mixture in $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ — $4$ g of hydrogen with $16$ g of oxygen — the oxygen is the reagent that runs out. Work out how much water that allows.

20. Practice

$46$ g of sodium is mixed with $142$ g of chlorine and the reaction $\mathrm{2Na + Cl_2 \rightarrow 2NaCl}$ goes to completion. What mass of sodium chloride is formed? Answer in grams.

21. Practice

Railroad crews weld steel rails with thermite: $\mathrm{Fe_2O_3 + 2Al \rightarrow Al_2O_3 + 2Fe}$. A charge holds $800$ g of iron oxide (160 g/mol) and $405$ g of aluminum (27 g/mol). What mass of molten iron, in grams, can it make?

22. Somewhere new

A plant runs $\mathrm{Zn + 2HCl \rightarrow ZnCl_2 + H_2}$ and charges each reactor with $715$ kg of zinc and $401.5$ kg of hydrogen chloride — the same proportions as the bench mixture, $11$ times larger and in kilograms. What mass of hydrogen does one batch make? Answer in kilograms.

23. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

24. Test question

The reaction is $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$, and someone has weighed out $4$ g of hydrogen and $16$ g of oxygen. Work out, for each reagent, how many moles there are and what that is divided by its coefficient.

25. What you can do now

You can find the limiting reagent and the mass of product it allows. Say which runs out first when 4 g of hydrogen meets 16 g of oxygen, and why the answer is the heavier of the two. Next: the difference between the mass the equation allows and the mass that actually comes out of the flask.

Working for the steps left to you

	amount formed, in mol	mass formed, in g
$\mathrm{H_2O}$

	mass weighed out, in g	amount, in mol	amount divided by the coefficient
$\mathrm{H_2}$	4
$\mathrm{O_2}$	16

16. Your turn: $\mathrm{2Na + Cl_2 \rightarrow 2NaCl}$, with 46 g of sodium and 142 g of chlorine., step 3

$1 \times 2 \times 58.5 = 117 \text{ g}$

Sodium limits; the reaction runs once.