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Wize University Chemistry Textbook > Stoichiometry

Limiting Reagents

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Introduction to Limiting Reagents

Anytime reactant species are in limited supply and not present in perfectly proportional amounts, a chemical reaction will have a limiting reagent which will be totally consumed before any other reactant

The quantity of the limiting reagent available directly determines the maximum number of product molecules that can be formed!

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Let's Consider an Example:

When making smores (yum!!) the "reaction" looks something like:



2 graham crackers + 1 piece of chocolate + 1 marshmellow → 1 smore


If I had 10 graham crackers, 6 pieces of chocolate, and 6 marshmellows, what would be the limiting reagent? In other words, what would I run out of first? And how many smores could I make?
  • The "chemistry way" to figure this out would be to take the # of moles of each reactant and divide by its stoichiometric coefficient:

Graham crackers: 10/2=5
Pieces of chocolate: 6/1=6
Marshmellows: 6/1=6

  • Now, to figure out the limiting reagent, look at which of the above numbers are the smallest!!
  • 5 is smallest, therefore Graham crackers are the limiting reagent!!

Now how many smores could we make?

We know graham crackers will determine how much product we get since we'll run out of the crackers first.

moles of graham cracker×1 mol smores2 moles graham cracker= moles of smoresmoles\ of\ graham\ cracker\times\frac{1\ mol\ smores}{2\ moles\ graham\ cracker}=\ moles\ of\ smores

10 x (1/2)= 5 moles of smores created!

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Wize Concept
It is necessary to determine the limiting reagent whenever we are given the amounts of two or more reactants in a chemical reaction.

The limiting reagent is totally consumed in the reaction. Any other reactant is in excess.

Limiting reagent stops the reaction by running out first.

The quantity of the limiting reagent available directly determines the maximum number of products molecules that can be formed


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Example: Determine the Mass of Product in a Limiting Reagent Problem

Iron and chlorine gas react to form iron (III) trichloride. If 110 g of iron and 105 g of chlorine gas are reacted, which species is the limiting reagent? What is the maximum mass of FeCl3\text{FeCl}_3that can be formed?

2Fe(s) + 3Cl2(g) → 2FeCl3(s)


Wize Tip
Steps for Solving Limiting Reagent Problems:

Step 1 – Write & balance the equation.
Step 2 – Calculate moles of each reactant.
Step 3 – Use the molar ratios of the present reactants (from the balanced equation) to determine the limiting reactant (LR).
  • Take the # of moles of each reactant from step 2 and divide by the stoichiometric coefficient for that reactant
  • Smallest value from this calculation tells us the LR
Step 4 – From the limiting reactant, use the molar ratio (from the balanced equation) to calculate moles of the desired product.
Step 5 – Convert moles to the desired units (density, molarity, grams, etc…)



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2Fe(s) + 3Cl2(g) → 2FeCl3(s)

Step 1 – Write & balance the equation.

Is the equation given balanced?
Yes


Step 2 – Calculate moles of each reactant.

i) Find moles of Fe

n=m/M
n=110/55.845
n=1.9697 moles


ii) Find moles of Cl2

n=m/M
n=105/70.906
n=1.48083 moles


Step 3 – Use the molar ratios of the present reactants (from the balanced equation) to determine the limiting reactant.

Divide # of moles of Fe by 2: 1.9697/2 =0.98495

Divide # of moles of Cl2 by 3: 1.48083/3=0.49361


Therefore the limiting reagent is
Cl2
! (because it had the smallest number)

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Step 4 – From the limiting reactant, use the molar ratio (from the balanced equation) to calculated moles of the desired product.

  • The maximum mass of FeCl3 that is formed depends on how much
    Cl2
    we have (LR).
  • Let's use moles of Cl2 to find moles of FeCl3

number of moles of Cl2 x (2 moles FeCl3)3 moles Cl2= number of  moles of FeCl3number\ of\ moles\ of\ Cl_2\ x\ \frac{\left(2\ moles\ FeCl_3\right)}{3\ moles\ Cl_2}=\ number\ of\ \ moles\ of\ FeCl_3

We found n=1.48083 for Cl2 --> (1.48083 x 2/3 =0.98722)
moles of FeCl3=0.98722 moles



Step 5 – Convert moles to the desired units (density, molarity, grams, etc…)


m(FeCl3)=?
M (FeCl3) =162.2g/mol
n=m/M
m=nM
m=0.98722mol x 162.2g/mol
m=160.13g
Therefore, 160g of FeCl3 will be formed from the reaction!

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Example: What is the Limiting Reagent?

2.0 g of aqueous Barium hydroxide (Ba(OH)2(aq), 171.32 g/mol) and 1.5 g of liquid hydrogen bromide (HBr(l), 80.91g/mol) react together to give aqueous barium bromide (BaBr2(aq)) and liquid water (H2O(l)). Which species would be the limiting reagent of this reaction?

Ba(OH)2(aq)+2HBr(ℓ)→BaBr2(aq)+2H2O(ℓ)Ba\left(OH\right)_2\left(aq\right)+2HBr\left(\ell\right)\rightarrow BaBr_2\left(aq\right)+2H_2O\left(\ell\right)


1) Calculate moles of each reactant

i) moles of Ba(OH)2:
n=m/M
n=2g/171.32g/mol
n=0.011674 moles

ii) moles of HBr:
n=m/M
n=1.5g/80.91g/mol
n=0.01854 moles

2) Now divide the # of moles we just calculated by the stoichiometric coefficients in the balanced equation:

i) Ba(OH)2
0.011674/1=0.011674

ii) HBr
0.01854/2=0.00927

This shows that HBr is the limiting reagent (we got the smallest number for it)
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Example: How Many Grams of the Excess Reagent Will Remain?


The reaction between P4 and Br2 is very exothermic and produces PBr5 as the only product. If 7.0 g of P4 react with 12.0 g of Br2 how many grams of the excess reagent will remain?

a) 0.4 g
b) 4.7 g
c) 6.1g
d)11.1 g

First let's start by writing the reaction out that was described above and balancing it. Don't worry if you didn't include the phases.
P4(s)+10 Br2(l)→4 PBr5(s)P_{4(s)}+10\ Br_{2(l)}\rightarrow 4\ PBr_{5(s)}
Now let's find the moles of each reactant,


To determine the LR, divide the moles by the stoichiometry coefficients:
i) P4:
0.0565/1=0.0565

ii) Br2:
0.0751/10=0.00751

The smallest number tells us the limiting reagent. The LR is Br2.
The LR (Br2) will be completely consumed.

What amount of P4 will react with Br2?

0.0751mol Br2 x (1 mol P4)(10 mol Br2)=0.00751 mol P40.0751mol\ Br_2\ x\ \frac{\left(1\ mol\ P_4\right)}{\left(10\ mol\ Br_2\right)}=0.00751\ mol\ P_4

Now that we calculated the amount of P4 that would completely react with Br2, we can determine the amount of P4 that would be left over.
nP4,remaining=nP4,starting−nP4,consumed=0.0565 mol−0.00751 mol=0.0490 moln_{P_4, remaining}=n_{P_4, starting}-n_{P_4, consumed}=0.0565\ mol-0.00751\ mol=0.0490\ mol

We can now convert that number of moles to mass and answer the problem,
n=m/M
m=nm
mP4,remaining=nP4,remaining×MW=0.0490 mol×123.90 g/mol=6.07g\displaystyle m_{P_4,remaining}=n_{P_4,remaining}\times MW=0.0490\ mol\times123.90\ g/mol=6.07g

Practice: What is the Limiting Reagent

If 12 g of sodium is mixed with 0.64 g of H2 and 17 g of O2 which reagent will be limiting in the reaction shown below?

2 Na(s)+O2(g)+H2(g)→2 NaOH2\ Na_{(s)}+O_{2(g)}+H_{2(g)}\rightarrow 2\ NaOH


Extra Practice
Stoichiometry and Limiting Reagent
Consider the following balanced equation:
2 NH3 (g) + 3 CuO (s) → N2 (g) + 3 Cu (s) + 3 H2O (l)
A reaction mixture contains 18.1 g NH3 and 90.4 g CuO. How much of the limiting reagent in (g's) is remaining?
Chemical reactions: Limiting Reagent
Sodium metal reacts with water to give sodium hydroxide and hydrogen gas as per the following unbalanced equation:
Na(s)+H2O(l)→NaOH(aq)+H2(g)Na(s)+H_2O(l)\rightarrow NaOH(aq)+H_2(g)
If 1.00 g of sodium were reacted with 0.786 g of water which species would be the limiting reagent?
Stoichiometry and Limiting Reagent
Consider the following balanced equation:
2 NH3 (g) + 3 CuO (s) → N2 (g) + 3 Cu (s) + 3 H2O (l)
A reaction mixture contains 18.1 g NH3 and 90.4 g CuO. How much of the limiting reagent in (g's) is remaining?
Stoichiometry: Limiting Reagents
Vanillin acetate is a synthetic flavoring agent used to mimic the natural flavor of vanilla in candies, beverages, ice creams, and baked goods. You only need 10-30 ppm concentrations of vanillin acetate to flavor most foods! The preparation of this compound is described by the balanced reaction scheme below. Use the data provided to answer the following questions.
Chemical reactions: Limiting Reagent & Percent Yield
Iron oxidizes when exposed to oxygen to form iron oxide according to the following equation:
Fe (s) + O2 (g) ​→ Fe2​O3​ (s)
352g of pure iron is exposed to 12.0 mols of O2, and after a period of time, 46.7 g of iron oxide (rust) is collected.
Sodium metal reacts with water to give sodium hydroxide and hydrogen gas as per the following unbalanced equation:
Na(s)+H2O(l)→NaOH(aq)+H2(g)Na(s)+H_2O(l)\rightarrow NaOH(aq)+H_2(g)
If 1.00 g of sodium were reacted with 0.786 g of water which species would be the limiting reagent?
Gas Stoichiometry and Limiting Reagent
Consider the following equation:
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(l)
3 L NH3 and 7 L O2 were reacted at 350 °C and 111 kPa. How much NO2 can be produced?
Stoichiometry: Yield
What is the maximum amount of iodine gas (in moles) that can be formed if 1.00 mole of manganese (IV) oxide was allowed to react with 2.00 moles of hydroiodic acid to from manganese(II) iodide, iodine gas, and water?
Chemical reactions: Limiting Reagent & Percent Yield
Iron oxidizes when exposed to oxygen to form iron oxide according to the following equation:
Fe (s) + O2 (g) ​→ Fe2​O3​ (s)
352g of pure iron is exposed to 12.0 mols of O2, and after a period of time, 46.7 g of iron oxide (rust) is collected.
Stoichiometry: Limiting reagents
3. Consider the chemical reaction described below:
Ca3(PO4)2 (s)    +    C (s)    +    SiO2 (s)    →    CaSiO3 (s)    +    CO (g)    +    P (l)Ca_3\left(PO_4\right)_2\ \left(s\right)\ \ \ \ +\ \ \ \ C\ \left(s\right)\ \ \ \ +\ \ \ \ SiO_2\ \left(s\right)\ \ \ \ \rightarrow\ \ \ \ CaSiO_3\ \left(s\right)\ \ \ \ +\ \ \ \ CO\ \left(g\right)\ \ \ \ +\ \ \ \ P\ \left(l\right)
a) What is the limiting reagent for this reaction if an initial reaction mixture contains 1325 kg of calcium phosphate, 300 kg of carbon and 890 kg of silicon dioxide.
Chemical Reactions: Limiting Reagent & Percent Yield Practice
Iron oxidizes when exposed to oxygen to form iron oxide according to the following equation:
Fe + O2 ​→Fe2​O3​
352g of pure iron is exposed to 12.0 mols of O2, and after a period of time 46.7 g of iron oxide (rust) is collected.
Stoichiometry: Limiting reagents
Ammonia, NH3, and borane, BH3, react to form a Lewis adduct, ammonia borane as shown below.
NH3(sol)+BH3(sol)→NH3BH3(s)NH_{3(sol)} + BH_{3(sol)} \rightarrow NH_3BH_{3(s)}
If 24 mL of a 0.74 M solution of ammonia is mixed with 12 mL of a 1.8 M solution of borane what is the maximum mass of ammonia borane that can be produced?
Stoichiometry: Limiting Reagents
When we are solving a stoichiometry problem and are given a balanced reaction, the limiting reagent is the reagent that...
(There are 2 correct answers)
Stoichiometry: Limiting reagents
17g of P4 reacts with 64g of oxygen, O2 as described below in the UNBALANCED reaction. What is the maximum amount of P4O6 that can be formed?
P4(s)+O2(g)→P4O6(s)P_{4(s)}+O_{2(g)}\rightarrow P_4O_{6(s)}
Chemical Reaction: Limiting Reagent
If 12 g of sodium is mixed with 0.64 g of H2 and 17 g of O2 which reagent will be limiting in the reaction shown below?
2 Na(s)+O2(g)+H2(g)→2 NaOH2\ Na_{(s)}+O_{2(g)}+H_{2(g)}\rightarrow 2\ NaOH
Chemical Reactions: Limiting Reagents
Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor:
4NH3+5O2→4NO+6H2O\rm 4NH_3+5O_2\rightarrow4NO+6H_2O
When 20.0 g NH3\rm NH_3 and 50.0 g O2\rm O_2 are allowed to react, which is the limiting reagent?
Chemical Reactions: Limiting Reagents
Magnesium metal reacts with HCl\text{HCl} to produce magnesium chloride (MgCl2\text{MgCl}_ 2 ) and hydrogen gas (H2\text{H}_2 ). If 0.568 g of magnesium are reacted with 75.0 mL of a 0.450 M solution of HCl\text{HCl}, how much magnesium metal will remain after the reaction is complete?
Chemical Reactions: Limiting Reagent & Percent Yield Practice
Iron oxidizes when exposed to oxygen to form iron oxide according to the following equation:
Fe + O2 ​→Fe2​O3​
352g of pure iron is exposed to 12.0 mols of O2, and after a period of time 46.7 g of iron oxide (rust) is collected.
Chemical Reactions: Limiting Reagents
Consider the following unbalanced reaction
Which reagent is in excess?
Chemical reactions: Limiting Reagent
Sodium metal reacts with water to give sodium hydroxide and hydrogen gas as per the following unbalanced equation:
Na(s)+H2O(l)→NaOH(aq)+H2(g)Na(s)+H_2O(l)\rightarrow NaOH(aq)+H_2(g)
If 1.00 g of sodium were reacted with 0.786 g of water which species would be the limiting reagent?