When there is not enough of one reactant in a inter-base.netical reaction, the reaction stops abruptly. To figure out the amount of product produced, it must be determined reactant will limit the inter-base.netical reaction (the limiting reagent) and which reactant is in excess (the excess reagent). One way of finding the limiting reagent is by calculating the amount of product that can be formed by each reactant; the one that produces less product is the limiting reagent.

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Introduction

The following scenario illustrates the significance of limiting reagents. 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 be made? With 20 tires, 5 cars can be produced because there are 4 tires to a car. With 14 headlights, 7 cars can be built (each car needs 2 headlights). Although more cars can be made from the headlights available, only 5 full cars are possible because of the limited number of tires available. 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). This scenario is illustrated below:

4 Tires + 2 Headlights = 1 Car

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

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

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

There are two ways to determine the limiting reagent. One method is to find and compare the mole ratio of the 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 smallest amount of product is the limiting reagent (approach 2).



How to Find the Limiting Reagent: Approach 2

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

Balance the inter-base.netical equation for the inter-base.netical reaction. Convert the given information into moles. Use stoichiometry for each individual reactant to find the mass of product produced. The reactant that produces a lesser amount of product is the limiting reagent. The reactant that produces a larger amount of product is the excess reagent. To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given.

b. If all of the 0.1388 moles of glucose were used up, there would need to be 0.1388 x 6 or 0.8328 moles of oxygen. Because there is an excess of oxygen, the glucose amount is used to calculate the amount of the products in the reaction.
Therefore, the mole ratio is: (0.8328 mol O2)/(0.208 mol C6H12O6)

This gives 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: (mathrm0.1388: moles: glucose imes dfrac61 = 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




Step 5: The reactant that produces a larger amount of product is the excess reagent
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.

(mathrm2.40:g: Mg imes dfrac1.00: mol: Mg24.31:g: Mg imes dfrac1.00: mol: O_22.00: mol: Mg imes dfrac32.0:g: O_21.00: mol: O_2 = 1.58:g: O_2) OR Mass of excess reagent calculated using the mass of the product: (mathrm3.98:g: MgO imes dfrac1.00: mol: MgO40.31:g: MgO imes dfrac1.00: mol: O_22.00: mol: MgO imes dfrac32.0:g: O_21.00: mol: O_2 = 1.58:g: O_2) Mass of total excess reagent given – mass of excess reagent consumed in the reaction10.0g – 1.58g = 8.42g O2 is in excess.

Example (PageIndex3): Limiting Reagent

What is the limiting reagent if 76.4 grams of (C_2H_3Br_3) were reacted with 49.1 grams of (O_2)?

Solution

Using Approach 1:

A. (mathrm76.4:g imes dfrac1: mole266.72:g = 0.286: moles: of: C_2H_3Br_3)

(mathrm49.1: g imes dfrac1: mole32:g = 1.53: moles: of: O_2)

B. Assuming that all of the oxygen is used up, (mathrm1.53 imes dfrac411) or 0.556 moles of C2H3Br3 are required. Because there are only 0.286 moles of C2H3Br3 available, C2H3Br3 is the limiting reagent.

Using Approach 2:

(mathrm76.4:g: C_2H_3Br_3 imes dfrac1: mol: C_2H_3Br_3266.72:g: C_2H_3Br_3 imes dfrac8: mol: CO_24: mol: C_2H_3Br_3 imes dfrac44.01:g: CO_21: mol: CO_2 = 25.2:g: CO_2)

(mathrm49.1:g: O_2 imes dfrac1: mol: O_232:g: O_2 imes dfrac8: mol: CO_211: mol: O_2 imes dfrac44.01:g: CO_21: mol: CO_2 = 49.1:g: CO_2)

Therefore, by either method, C2H3Br3is the limiting reagent.


A. (mathrm78:g imes dfrac1: mol77.96:g = 1.001: moles: of: Na_2O_2)

(mathrm29.4:g imes dfrac1: mol18:g= 1.633: moles: of: H_2O)

B. Assume that all of the water is consumed, (mathrm1.633 imes dfrac22) or 1.633 moles of Na2O2 are required. Because there are only 1.001 moles of Na2O2, it is the limiting reactant.

Using Approach 2:

(mathrm78:g: Na_2O_2 imes dfrac1: mol: Na_2O_277.96:g: Na_2O_2 imes dfrac4: mol: NaOH2: mol: Na_2O_2 imes dfrac40:g: NaOH1: mol: NaOH = 80.04:g: NaOH)

Using either approach gives Na2O2 as the limiting reagent.

Example (PageIndex6): Identifying the Limiting Reagent

Will 28.7 grams of (SiO_2) react completely with 22.6 grams of (H_2F_2)? If not, identify the limiting reagent.

Solution

A. (mathrm28.7:g imes dfrac1: mole60.08:g = 0.478: moles: of: SiO_2)

(mathrm22.6:g imes dfrac1: mole39.8:g = 0.568: moles: of: H_2F_2)

B. There must be 1 mole of SiO2 for every 2 moles of H2F2 consumed. Because the ratio is 0.478 to 0.568, 28.7 grams of SiO2 do not react with the H2F2.

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C. Assuming that all of the silicon dioxide is used up, (mathrm0.478 imes dfrac21) or 0.956 moles of H2F2 are required. Because there are only 0.568 moles of H2F2, it is the limiting reagent.