Limiting Reagent Examples

Limiting Reagent Problems #1-10 | Limiting Reagent Problems #11-20 | Stoichiometry Menu |

**First comment before starting:**

Just a bit below, I'm going to tell you (several times) how to determine the limiting reagent in a chemistry problem. I certainly hope it is something you pay attention to and remember. Figuring out which substance is the limiting reagent is an area that many students struggle with.

You will see the word "excess" used in this section and in the problems. It is used several different ways:

(a)Compound A reacts with an excess of compound B.In this case, mentally set compound B aside for the moment. Since it is "in excess," this means there is more than enough of it. The other compound will run out first.(b)

20 grams of A and 20 grams of B react. Which is in excess?What we will do below is find out which substance runs out first (called the limiting reagent). Obviously (I hope), the other compound is seen to be in excess.(c)

After 20 gm. of A and 20 gm. of B react, how much of the excess compound remains.To answer this problem, a subtraction will be involved. This is a part of many limiting reagent problems and it causes difficult with students. Expect it to be on your test.

**Second comment before starting: What is the Limiting Reagent?**

It is simply the substance in a chemical reaction that runs out first. It seems to be a simple concept, but it does cause people problems. Let's try a simple non-chemical example.

Reactant A is a test tube. I have 20 of them.

Reactant B is a stopper. I have 30 of them.

Product C is a stoppered test tube.

The reaction is:

A + B ---> Ctest tube plus stopper gives stoppered test tube.

So now we let them "react." The first stopper goes in, the second goes in and so on. Step by step we use up stoppers and test tubes (the amounts go down) and make stoppered test tubes (the amount goes up).

Suddenly, we run out of one of the "reactants." Which one? We run out of test tubes first. Seems obvious, doesn't it? We had 20 test tubes, but we had 30 stoppers. So when the test tubes are used up, we have 10 stoppers sitting there unused. And we also have 20 test tubes with stoppers firmly inserted.

So, which "reactant" is limiting and which is in excess? The test tubes are limiting (they ran out first) and the stoppers are in excess (we have some left over when the limiting reagent ran out).

There are two techniques for determine the limiting reagent in chemical problems. The first technique is discussed as part of the solution to the first example. Make sure you take a close look at it. The second technique will make its first appearance in Example #6.

**Example #1:** Here's a nice limiting reagent problem we will use for discussion. Consider the reaction:

2Al + 3I_{2}------> 2AlI_{3}

Determine the limiting reagent and the theoretical yield of the product if one starts with:

(a) 1.20 mol Al and 2.40 mol iodine.

(b) 1.20 g Al and 2.40 g iodine

(c) How many grams of Al are left over in part b?

**Solution for part (a):**

We already have moles as the unit, so we use those numbers directly.

1) Here is how to find out the limiting reagent:

take the moles of each substance and divide it by its coefficient in the balanced equation. The substance that has the smallest answer is the limiting reagent.

2) Let's say that again:

to find the limiting reagent, take the moles of each substance and divide it by its coefficient in the balanced equation. The substance that has the smallest answer is the limiting reagent.You're going to need that technique, so remember it.

By the way, did you notice that I bolded the technique to find the limiting reagent? I did this so as to emphasize its importance to you when learning how to do limiting reagent problems.

3) Resuming with the problem solution:

For aluminum: 1.20 / 2 = 0.60

For iodine: 2.40 / 3 = 0.80

4) The lowest number indicates the limiting reagent. Aluminum will run out first in part (a) of the question. Why?

1.20/2 means there are 0.60 "groupings" of 2 and 2.40/3 means there are 0.80 "groupings" of 3. If they ran out at the same time, we'd need one "grouping" of each. Since there is less of the "grouping of 2," it will run out first.If you're not sure what I just said, that's OK. The technique works, so remember it and use it.

5) The second part of the question "theoretical yield" depends on finding out the limiting reagent. Once we do that, it becomes a stoichiometric calculation.

Al and AlI_{3}stand in a one-to-one molar relationship, so 1.20 mol of Al produces 1.20 mol of AlI_{3}. Notice that the amount of I_{2}does not play a role, since it is in excess.

**Solution for part (b):**

1) Since we have grams, we must first convert to moles. The we solve just as we did in part a just above. For the mole calculation:

aluminum is 1.20 g / 26.98 g mol¯^{1}= 0.04477 mol

iodine is 2.4 g / 253.8 g mol¯^{1}= 0.009456 mol

2) To determine the limiting reagent:

aluminum is 0.04477 / 2 = 0.02238

iodine is 0.009456 / 3 = 0.003152The lower number is iodine, so we have identified the limiting reagent.

3) Finally, we have to do a calculation and it will involve the iodine, NOT the aluminum.

I_{2}and AlI_{3}stand in a three-to-two molar relationship, so 0.009456 mol of I_{2}produces 0.006304 mol of AlI_{3}. Again, notice that the amount of Al does not play a role, since it is in excess.From here figure out the grams of AlI

_{3}and you have your answer.

**Solution for part (c):**

Since we have moles, we calculate directly and then convert to grams.Al and I

_{2}stand in a two-to-three molar relationship, so 0.009456 mol of I_{2}uses 0.006304 mol of Al.Convert this aluminum amount to grams and subtract it from 1.20 g and that's the answer.

Just above was some discussion on a way to determine the limiting reagent in a chemistry problem. This particular thing (determine the limiting reagent) is a real stumbling block for students. Be aware!

**Example #2:** 15.00 g aluminum sulfide and 10.00 g water react until the limiting reagent is used up. Here is the balanced equation for the reaction:

Al_{2}S_{3}+ 6H_{2}O ---> 2Al(OH)_{3}+ 3H_{2}S(a) Which is the limiting reagent?

(b) What is the maximum mass of H_{2}S which can be formed from these reagents?

(c) How much excess reagent remains after the reaction is complete?

Some comments first:

The key to this problem is the limiting reagent, part (a). Once you know that, part (b) becomes "How much H_{2}S can be made from the limiting reagent?" Part (c) becomes two connected questions: first, "How much Al_{2}S_{3}is used up when reacting with the limiting reagent?" then second, "What is 15.00 minus the amount in the first part?"Make sure you note that second part. The calculation gives you the answer to "How much reacted?" but the question is "How much remained?" Lots of students forget to do the second part (the 15 minus part) and so get graded down.

Note: I'm carrying a guard digit or two through the calculations. The final answers will appear with the proper number of significant figures.

**Solution for limiting reagent, part (a):**

1) Determine the moles of Al_{2}S_{3} and H_{2}O

aluminum sulfide: 15.00 g ÷ 150.158 g/mol = 0.099895 mol

water: 10.00 g ÷ 18.015 g/mol = 0.555093 mol

2) Divide each mole amount by equation coefficient

aluminum sulfide: 0.099895 mol ÷ 1 mol = 0.099895

water: 0.555093 mol ÷ 6 mol = 0.0925155

3) The water is the lesser amount; it is the limiting reagent.

**Solution for mass of H _{2}S formed, part (b)**

Now that we know the limiting reagent is water, this problem becomes "How much H_{2}S is produced from 10.00 g of H_{2}O and excess aluminum sulfide?"

1) Determine moles of 10.00 g of H_{2}O

water: 10.00 g ÷ 18.015 g/mol = 0.555093 mol

2) Use molar ratios to determine moles of H_{2}S produced from above amount of water.

(a) the H_{2}O/H_{2}S molar ratio is 6/3, a 2/1 ratio.

(b) water is associated with the two. This means the H_{2}S amount is one-half the water value = 0.2775465 mol.

3) Convert moles of H_{2}S to grams.

0.2775465 mol x 34.0809 g/mol = 9.459 g

**Solution for excess reagent remaining, part (c)**

We will use the amount of water to calculate how much Al_{2}S_{3} reacts, then subtract that amount from 15.00 g.

1) Determine moles of 10.00 g of H_{2}O

water: 10.00 g ÷ 18.015 g/mol = 0.555093 mol

2) Use molar ratios to determine moles of Al_{2}S_{3} that reacts with the above amount of water.

(a) the Al_{2}S_{3}/H_{2}O ratio is 1/6

(b) water is associated with the 6. This means the Al_{2}S_{3}amount is one-sixth the water value = 0.09251447 mol

3) Convert moles of Al_{2}S_{3} to grams.

0.09251447 mol x 150.158 g/mol = 13.891943 g

4) However, we are not done. We were asked for the amount remaining and the answer just above is the amount which was used up, so the final step is:

15.00 g - 13.891943 g = 1.108 g

**Example #3:** If there is 35.0 grams of C_{6}H_{10} and 45.0 grams of O_{2}, how many grams of the excess reagent will remain after the reaction ceases?

2C_{6}H_{10}+ 17O_{2}---> 12CO_{2}+ 10H_{2}O

**Solution:**

1) Convert each substance to moles:

C_{6}H_{10}: 35.0 g / 82.145 g/mol = 0.426 mol

O_{2}: 45.0 g / 31.998 g/mol = 1.406 mol

2) Determine the limiting reagent:

C_{6}H_{10}: 0.426 mol / 2 = 0.213

O_{2}: 1.406 mol / 17 = 0.083O

_{2}is the limiting reagent.Comment: the units don't matter in this step. What we are looking for is the smallest number after carrying out the divisions. The value of 0.083 is the important thing. Not if it has a unit attached to it or not.

3) Determine how many moles of the excess reagent is used up when the limiting reagent is fully consumed:

the mole ratio we desire is 2/17 (C_{6}H_{10}to O_{2})

2 x –––– = ––––––– 17 1.406 mol x = 0.1654 mol of C

_{6}H_{10}consumed

4) Determine grams of C_{6}H_{10} remaining:

0.426 mol minus 0.1654 mol = 0.2606 mol of C_{6}H_{10}remaining0.2606 mol times 82.145 g/mol = 21.4 g remaining (to three sig figs)

**Example #4:** (a) What mass of Al_{2}O_{3} can be produced from the reaction of 10.0 g of Al and 19.0 g of O_{3}? (b) How much of the excess reagent remains unreacted?

**Solution to a:**

1) Write balanced chemical equation:

2Al + O_{3}---> Al_{2}O_{3}

2) Convert grams to moles:

Al ---> 10.0 g / 26.982 g/mol = 0.37062 mol

O_{3}---> 19.0 g / 47.997 g/mol = 0.39586 mol

3) Determine limiting reagent:

Al ---> 0.37062 / 2 = 0.18531

O_{3}---> 0.39586 / 1 = 0.39586Al is the limiting reagent

4) Determine moles of product formed:

Al to Al_{2}O_{3}molar ratio is 2 to 1.

2 0.37062 mol –––– = –––––––––– 1 x x = 0.18531 mol

5) Determine grams of product:

0.18531 mol times 101.961 g/mol = 18.8944 gTo three sig figs, 18.9 g

**Solution to b:**

1) Determine moles of ozone that reacted:

Al to O_{3}molar ratio is 2 to 1

2 0.37062 mol –––– = –––––––––– 1 x x = 0.18531 mol

2) Determine moles of ozone remaining:

0.39586 mol - 0.18531 mol = 0.21055 mol

3) Determine grams of ozone remaining:

0.21055 mol times 47.997 g/mol = 10.1 g (to three sig figs)

**Example #5:** Based on the balanced equation:

C_{4}H_{8}+ 6O_{2}---> 4CO_{2}+ 4H_{2}O

Calculate the number of excess reagent units remaining when 28 C_{4}H_{8} molecules and 228 O_{2} molecules react?

**Solution:**

Remember, numbers of molecules are just like moles, so treating the 28 and 228 as moles is perfectly acceptable. This is because I could divide the 28 and the 228 by Avogadro's Number to obtain the moles. Those mole amounts could be used in the calculation below and the final answer could then be multiplied by Avogadro's Number to obtain the answer of 60.

1) Determine the limiting reagent:

butane: 28 / 1 = 28

oxygen: 228 / 6 = 38Butane is the limiting reagent.

2) Determine how much oxygen reacts with 28 C_{4}H_{8} molecules:

the butane:oxygen molar ratio is 1:628 x 6 = 168 oxygen molecules react

3) Determine excess oxygen:

228 - 168 = 60

Here's aother way to consider this:

The 38 above means that there are 38 "groupings" of six oxygen molecules.38 minus 28 = 10 oxygen "groupings" remain after the butane is used up

10 x 6 = 60

**Example #6:** Determine the maximum mass of TiCl_{4} that can be obtained from 35.0 g of TiO_{2}, 45.0 g Cl_{2} and 11.0 g of C. (See comment below problem.)

3TiO_{2}+ 4C + 6Cl_{2}---> 3TiCl_{4}+ 2CO_{2}+ 2CO

**Solution:**

1) Assume each reactant is the limiting reagent. Determine the moles of product produced by each assumption:

Note: the first factor in each case converts grams of each reactant to moles. The second factor uses a molar ratio from the chemical equation to convert from moles of the reactant to moles of product. There is no need to convert to grams because all three calculations yield moles of the same compound (the TiCl_{4}).

1 mole Cl _{2}3 mole TiCl _{4}45.0 g Cl _{2}x––––––––––– x ––––––––––– = 0.31732 mol TiCl _{4}70.9064 g Cl _{2}6 mol Cl _{2}

1 mole C 3 mole TiCl _{4}11.0 g C x ––––––––––– x ––––––––––– = 0.68688 mol TiCl _{4}12.01078 g C 4 mol C

1 mole TiO _{2}3 mole TiCl _{4}35.0 g TiO _{2}x––––––––––– x ––––––––––– = 0.438235 mol TiCl _{4}79.8658 g TiO _{2}3 mol TiO _{2}Cl

_{2}makes the least amount of TiCl_{4}, so Cl_{2}is the limiting reactant.

2) The mass of TiCl_{4} produced is:

(0.31732 mol TiCl_{4}) (189.679 g TiCl_{4}/mol) = 60.2 g TiCl_{4}(to three sig figs)

Note that the "divide moles by coefficient" was not used to determine the limiting reagent. Instead, a full calculation was done and the least amount of product identified the limiting reagent. Here is what the "divide moles by coefficient" set up looks like:

Cl_{2}---> 0.63464 / 6 = 0.10577 <--- there's our limiting reagent

C ---> 0.915844 / 4 = 0.228961

TiO_{2}---> 0.438235 / 3 = 0.14608

**Example #7:** Determine the starting mass of each reactant if 46.3 of K_{3}PO_{4} is produced and 92.8 of H_{3}PO_{4} remains unreacted.

3KOH(aq) + H_{3}PO_{4}(aq) ---> K_{3}PO_{4}(aq) + 3H_{2}O(ℓ)

**Solution:**

1) The fact that some phosphoric acid remains tells us it is the excess reagent. Let us determine the amount of KOH (the limiting reagent) required to produce the 46.3 g of K_{3}PO_{4}.

46.3 g / 212.264 g/mol = 0.2181246 mol of K_{3}PO_{4}Three moles of KOH are required to produce one mole of K

_{3}PO_{4}0.2181246 mol times 3 = 0.6543738 mol of KOH required

0.6543738 mol times 56.1049 g/mol = 36.7 g (to thee sig figs)

2) Determine the starting mass of H_{3}PO_{4}

0.2181246 mol of K_{3}PO_{4}requires 0.2181246 mol of H_{3}PO_{4}based on the 1:1 molar ratio from the balanced equation.0.2181246 mol times 97.9937 g/mol = 21.4 g (to three sig figs)

21.4 + 92.8 = 114.2 g

**Example #8:** Determine the limiting reagent of this reaction:

Na_{2}B_{4}O_{7}+ H_{2}SO_{4}+ 5H_{2}O ---> 4H_{3}BO_{3}+ Na_{2}SO_{4}There are 5.00 g of each reactant.

**Solution:**

1) Convert everything into moles, by dividing each 5.00 g by their respective molar masses:

Na_{2}B_{4}O_{7}---> 0.02485 mol

H_{2}SO_{4}---> 0.05097 mol

H_{2}O ---> 0.2775 mol

2) Note that there are three reactants. How is the limiting reagent determined when there are three reactants? Answer: determine the limiting reagent between the first two:

Na_{2}B_{4}O_{7}---> 0.02485 / 1 = 0.02485

H_{2}SO_{4}---> 0.05097 / 1 = 0.05097Na

_{2}B_{4}O_{7}is the limiting reagent when compared to H_{2}SO_{4}

3) Now, compare the "winner" to the third reagent:

Na_{2}B_{4}O_{7}---> 0.02485 / 1 = 0.02485

H_{2}O ---> 0.2775 / 5 = 0.0555Na

_{2}B_{4}O_{7}is the limiting reagent between itself and H_{2}O.Na

_{2}B_{4}O_{7}is the overall limiting reagent in this problem.

**Example #9:** How much O_{2} could be produced from 2.45 g of KO_{2} and 4.44 g of CO_{2}?

4KO_{2}+ 2CO_{2}---> 2K_{2}CO_{3}+ 3O_{2}

**Solution:**

I will do a solution assuming KO_{2} is the limiting reagent, then I will do a solution assuming CO_{2} is the limiting reagent. The reactant that produces the lesser amount of oxygen is the limiting reagent and that lesser amount will be the answer to the question.

1) Solution using KO_{2}:

2.45 g / 71.096 g/mol = 0.03446045 mol

4 0.03446045 mol ––– = –––––––––––––– 3 x x = 0.02584534 mol

(0.02584534 mol) (31.998 g/mol) = 0.827 g of O

_{2}

2) Solution using CO_{2}:

4.44 g / 44.009 g/mol = 0.10088845 mol

2 0.10088845 mol ––– = –––––––––––––– 3 x x = 0.151332 mol

(0.151332 mol) (31.998 g/mol) = 4.84 g of O

_{2}

3) 0.827 g is the answer.

Note that I could have calculated the mole amounts, used the "divide moles by coefficient" to determine the limiting reagent, and then done just one complete calculation.

**Example #10:** (a) What mass of hydrogen peroxide should result when 1.45 g of barium peroxide is treated with 25.5 mL of hydrochloric acid solution containing 0.0277 g of HCl per mL? (b) How much of the excess reactant is left?

BaO_{2}(s) + 2HCl(aq) ---> H_{2}O_{2}(aq) + BaCl_{2}(aq)

**Solution:**

Calculate the amount of product using each reactant. The reactant that produces the lesser of the two amounts will tell you the limiting reactant. This solution will use dimensional analysis (also called the unit-factor, or unit-label, method) for the proposed solution.

1) First, determine the mass of HCl that reacts:

(0.0277 g/mL) (25.5 mL) = 0.70635 g

2) The barium peroxide solution:

(1.45 g BaO _{2})(1 mol BaO _{2}/ 169.3 g BaO_{2})(1 mol H _{2}O_{2}/ 1 mol BaO_{2})(34.0 g H _{2}O_{2}/ 1 mol H_{2}O_{2})= 0.291 g H _{2}O_{2}↑ convert grams to moles ↑ ↑ molar ratio ↑

from equation↑ convert moles to grams ↑

3) The hydrochloric acid solution:

(0.70635 g) (1 mol HCl / 36.46 g HCl) (1 mol H_{2}O_{2}/ 2 mol HCl) (34.0 g H_{2}O_{2}/ 1 mol H_{2}O_{2}) = 0.332 g H_{2}O_{2}

4) Since 0.291 g is less than 0.332 g, the BaO_{2} is the limiting reactant.

5) The other method to determine the limiting reagent is to divide the moles of each reactant by their respective coefficient in the balanced equation:

BaO_{2}---> 1.45 g / 169.3 g/mol = 0.008565 mol

HCl ---> 0.70635 g / 36.46 g/mol = 0.01937 mol0.008565 / 1 = 0.008565

0.01937 / 2 = 0.009685BaO

_{2}(the 0.008565) is the lesser amount, so it is the limiting reagent.

6) To solve part (b), we observe that 0.008565 mol of BaO_{2} was used. Using a 1:2 molar ratio, we can determine the amount of HCl that was used:

1 0.008565 mol ––– = ––––––––––– 2 x x = 0.01713 mol of HCl used up in the reaction

7) Next, we subtract the amount used up from the total amount that was present:

0.01937 mol - 0.01713 mol = 0.00224 mol of HCl remains after reaction stops

8) Convert moles to grams:

(0.00224 mol) (36.46 g/mol) = 0.0817 g (to three sig figs)

Limiting Reagent Problems #1-10 | Limiting Reagent Problems #11-20 | Stoichiometry Menu |