# Limiting Reagents

What happens when there is not enough of one reactant in a chemical reaction? The reaction stops abruptly because there is too little of one reactant. How do you figure out the amount of product that is produced? To do so, we must first find which reactant will limit the chemical reaction (the limiting reagent) and which reactant will be in excess (the excess reagent). One way of finding the limiting reagent is by calculating the amount of product formed by each reactant; the one that produces less amount of product is the limiting reagent.

### Introduction

Let us now turn our attention to a real life problem that simulates the importance of limiting reagents. Suppose that you are in a car factory. In order to assemble a car, 4 tires and 2 headlights are needed (among other things). In this example, imagine that the tires and headlights are reactants while the car is the product formed from the reaction of 4 tires and 2 headlights. If you have 20 tires and 14 headlights, how many cars can you make? With 20 tires, you can produce 5 cars because there are 4 tires to a car. With 14 headlights, 7 cars can be built (each car needs 2 headlights). Although you can make more cars from the amount of headlights available, you can only create 5 full cars because of the number of tires available, making the tires the limiting reagent in this "reaction". In this case, the headlights are in excess. Because the number of cars formed by 20 tires is less than number of cars produced by 14 headlights, the tires are the limiting reagent (they limit the full completion of the reaction, in which all of the reactants are used up). We can visualize this as well:

4 Tires + 2 Headlights = 1 Car

Figure 1: The synthesis reaction of making a car. Images used from Wikipedia with permission.

The initial condition here is that there must be 4 tires to 2 headlights, thus the reactants must occur in that ratio; otherwise one will limit the reaction. We have 20 tires and 14 headlights, so there are two ways of looking at this problem. For 20 tires, we need 10 headlights while for 14 headlights, we need 28 tires. Since we do not have enough tires (20 tires is less than the 28 that we need), tires are the limiting "reactant".

### Definition

The limiting reagent is the reactant that is completely used up in a reaction and thus determines when the reaction stops. From stoichiometry, one can calculate the exact amount of reactant needed to react with another element. If the reactants are not mixed in the correct stoichiometric proportions (as seen in the balanced chemical equation), then one of the reactants will be entirely consumed while another will be left over in excess. The limiting reagent is the one that is totally consumed. The limiting reagent limits the reaction from continuing because there is none left to react with the now excess reactant.

### How to Find the Limiting Reagent

There are two main ways to determine the limiting reagent. One way is to find and compare the mole ratio of the amount of reactants used in the reaction (approach 1). Another way is to calculate the grams of products produced from the given quantities of reactants; the reactant that produces the least amount of product is the limiting reagent (approach 2).

Approach 1: Find the limiting reagent by looking at the number of moles of each reactant.

1. Determine the balanced chemical equation for the chemical reaction.
2. Convert all given information into moles (most likely, through the use of molar mass as a conversion factor).
3. Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
4. Use the amount of limiting reactant to calculate the amount of product produced.
5. If necessary, calculate how much is left in excess of the non-limiting reagent.

Approach 2: Find the limiting reagent by calculating and comparing the amount of product each reactant will produce.

1. Balance the chemical equation for the chemical reaction.
2. Convert the given information into moles.
3. Use stoichiometry for each individual reactant to find the mass of product produced.
4. The reactant that produces a lesser amount of product is the limiting reagent.
5. The reactant that produces a higher amount of product is the excess reagent.
6. To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given.

Example 1: Photosynthesis
Let us look at respiration, one of the most common chemical reactions on earth.

$C_6H_{12}O_6 + 6 O_2 \rightarrow 6 CO_2 + 6 H_2O + \rm{energy}$

What mass of carbon dioxide forms in the reaction of 25 grams of glucose with 40 grams of oxygen?

SOLUTION

When approaching this problem, you should be able to see that for every 1 mole of glucose ($$C_6H_{12}O_6$$), you need 6 moles of oxygen to obtain 6 moles of carbon dioxide and 6 moles of water.

Step 1: Determine the balanced chemical equation for the chemical reaction.

Step 2: Convert all given information into moles (most likely, through the use of molar mass as a conversion factor).

25g x (1 mol/180.06g) = 0.1388 mol  $$C_6H_{12}O_6$$

40g x (1 mol/32g) = 1.25 mol $$O_2$$

Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.

a. If all of the 1.25 moles of oxygen were to be used up, there would need to be 1.25 x (1/6) or 0.208 moles of glucose. There is only 0.1388 moles of glucose available which makes it the limiting reactant.

$1.25;\; \rm{mol} \; O_2 \times \dfrac{ 1 \; \rm{mol} \; C_6H_{12}O_6}{6\; \rm{mol} \; O_2}= 0.208 \; \rm{mol} \; C_6H_{12}O_6$

b. If all of the 0.1388 moles of glucose were to be used up, there would need to be 0.1388 x 6 or 0.8328 moles of oxygen. Since there is an excess of oxygen, the glucose amount will be used to calculate the amount of the products in the reaction.

$0.1388\; \rm{ mol}\; C_6H_{12}O_6 \times \dfrac{6 \; \rm{mol} \;O_2}{1 \; \rm{mol} \; C_6H_{12}O_6} = 0.8328 \; \rm{mol}\; O_2$

If more than 6 moles of Ois available per mole of C6H12O6, the oxygen is in excess and the glucose is the limiting reactant. If less than 6 mole of oxygen is available per mole of glucose, the oxygen is the limiting reactant. Our ratio is 6 mole oxygen per 1 mole glucose, OR 1 mole oxygen per 1/6 mole glucose. This means: 6 mol O/ 1 mol C6H12O6 .

Therefore the mole ratio is: (0.8328 mol O2)/(0.208 mol C6H12O6)

Which gives us a 4.004 ratio of O2 to C6H12O6.

Step 4: Use the amount of limiting reactant to calculate the amount of CO2 or H2O produced.

For carbon dioxide produced: 0.1388 moles glucose x (6/1) = 0.8328 moles carbon dioxide.

Step 5: If necessary, calculate how much is left in excess.

1.25 mol - 0.8328 mol = 0.4172 moles of oxygen left over

Example 2: Oxidation of Magnesium
Calculate the mass of magnesium oxide possible if 2.40 g $$Mg$$reacts with 10.0 g $$O_2$$

$Mg +O_2 \rightarrow MgO$

SOLUTION

Step 1: Balance equation

$2 Mg + O_2 \rightarrow 2 MgO$

Step 2 and Step 3: Converting mass to moles and stoichiometry

(2.40g Mg) x (1.00 mol Mg)/(24.31g Mg) x (2.00 mol MgO)/(2.00 mol Mg) x (40.31g MgO)/(1.00 mol MgO) = 3.98g MgO

(10.0g O2) x (1 mol O2)/(32.0g O2) x (2 mol MgO)/(1 mol O2) x (40.31g MgO)/(1 mol MgO) = 25.2g MgO

Step 4: The reactant that produces a lesser amount of product is the limiting reagent
Mg produces  less amount of MgO than O2 (3.98 g MgO vs. 25.2 g MgO), therefore Mg is the limiting reagent in this reaction.

Step 5: The reactant that produces a higher amount of product is the excess reagent
O2 produces more amount of MgO than Mg (25.2g MgO vs. 3.98 MgO), therefore O2 is the excess reagent in this reaction.

Step 6: Find the amount of remaining excess reactant by subtracting the mass of the excess reagent consumed from the total mass of excess  reagent given.
Mass of excess reagent calculated using the limiting reagent:
(2.40g Mg) x (1.00 mol Mg)/(24.31g Mg) x (1.00 mol O2)/(2.00 mol Mg) x (32.0g O2)/(1.00 mol O2) = 1.58g O
OR Mass of excess reagent calculated using the mass of the product:
(3.98g MgO) x (1.00 mol MgO)/(40.31g MgO) x (1.00 mol O2)/(2.00 mol MgO) x (32.0g O2)/(1.00 mol O2) = 1.58g O2
Mass of total excess reagent given – mass of excess reagent consumed in the reaction
10.0g – 1.58g = 8.42g O2 is in excess.
Example 3: Limiting Reagent

What would be the limiting reagent if 76.4 grams of $$C_2H_3Br_3$$ were reacted with 49.1 grams of $$O_2$$?

$4 C_2H_3Br_3 + 11 O_2 \rightarrow 8 CO_2 + 6 H_2O + 6 Br_2$

SOLUTION

Using Approach 1:

A. 76.4g x (1 mole/266.72g) = 0.286 moles of C2H3Br3

49.1g x (1 mole/32g) = 1.53 moles of O2

B. If we assume that all of the oxygen is used up, 1.53 x (4/11) or 0.556 moles of C2H3Br3 are required. Since there is only 0.286 moles of C2H3Br3 available, C2H3Br3 is the limiting reagent.

Using Approach 2:

76.4g C2H3Br3 x (1 mol C2H3Br3)/(266.72g C2H3Br3 ) x (8 mol CO2)/(4 mol C2H3Br3) x (44.01g CO2)/(1 mol CO2) = 25.2g CO2

49.1g Ox (1 mol O2)/(32g O2) x (8 mol CO2)/(11 mol O2) x (44.01g CO2)/(1 mol CO2) = 49.1g CO2

Therefore, we find that using either formulas, C2H3Br3is the limiting reagent.

Example 4: Limiting Reagent

What would be the limiting reagent if 78 grams of Na2O2 were reacted with 29.4 grams of H2O?

SOLUTION

Using Approach 1:

A. 78g x (1 mol/77.96g) = 1.001 moles of Na2O2

29.4g x (1 mol/18g)= 1.633 moles of H2O

B. If we assume that all of the water is used up, 1.633 x (2/2) or 1.633 moles of Na2O2 are required. Since there are only 1.001 moles of Na2O2, it is the limiting reactant.

Using Approach 2:

78g Na2O2 x (1 mol Na2O2)/(77.96g Na2O2) x (4 mol NaOH)/(2 mol Na2O2) x (40g NaOH)/(1 mol NaOH) = 80.04g NaOH

Therefore, we find that using either formulas yield $$Na_2O_2$$ as the limiting reagent.
Example 5: Excess Reagent

What is the amount of the excess reagent in the reaction where 24.5 grams of CoO is reacted with 2.58 grams of O2?

$4 CoO + O_2 \rightarrow 2 Co_2O_3$

SOLUTION

A. 24.5g x (1 mole/74.9g)= 0.327 moles of $$CoO$$

2.58g x (1 mole/32g)= 0.0806 moles of $$O_2$$

B. If we assume that all of the oxygen is used up, 0.0806 x (4/1) or 0.3225 of $$CoO$$ are required. Since there are 0.327 moles of CoO, CoO is in excess and thus O2 is the limiting reactant.

C. 0.327mol - 0.3224mol = 0.0046 moles left in excess.

Example 6: Identifying the Limiting Reagent

Will 28.7 grams of $$SiO_2$$ react completely with 22.6 grams of $$H_2F_2$$? If not, state which is the limiting reagent.

$SiO_2+ 2 H_2F_2 \rightarrow SiF_4+ 2 H_2O$

SOLUTION

A. 28.7g x (1 mole/60.08g) = 0.478 moles of $$SiO_2$$

22.6g x (1 mole/39.8g) = 0.568 moles of $$H_2F_2$$

B. There needs to be 1 mole of $$SiO_2$$ for 2 moles of $$H_2F_2$$ consumed. Since the ratio is 0.478 to 0.568, 28.7 grams of $$SiO_2$$ do not react with the $$H_2F_2$$.

C. If we assume that all of the silicone dioxide is used up, 0.478 x (2/1) or 0.956 moles of are $$H_2F_2$$required. Since there are only 0.568 moles of $$H_2F_2$$ it is the limiting reagent.

### References

1. Petrucci, Ralph H., William S. Harwood, Geoffery F. Herring, and Jeffry D. Madura. General Chemistry. 9th ed. New Jersey: Pearsin Prentice Hall, 2007.
2. Staley, Dennis. Prentice Hall Chemistry. Boston: Pearson Prentice Hall, 2007.

### Contributors

• Sarick Shah (UCD)

18:42, 21 Sep 2014

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