When tright here is not sufficient of one reactant in a jiyuushikan.orgical reactivity, the reactivity stops abruptly. To figure out the amount of product created, it should be determined reactant will certainly limit the jiyuushikan.orgical reactivity (the limiting reagent) and which reactant is in excess (the excess reagent). One method of finding the limiting reagent is by calculating the amount of product that have the right to be created by each reactant; the one that produces much less product is the limiting reagent.

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## Introduction

4 Tires + 2 Headlights = 1 Car

+
=
Figure 1: The synthesis reaction of making a automobile. Imperiods offered from Wikipedia with permission.

The initial problem is that there have to be 4 tires to 2 headlights. The reactants have to therefore take place in that ratio; otherwise, one will limit the reaction. Tright here are 20 tires and 14 headlights, so tbelow are 2 means of looking at this problem. For 20 tires, 10 headlights are compelled, whereas for 14 headlights, 28 tires are forced. Since tright here are not sufficient tires (20 tires is much less than the 28 required), tires are the limiting "reactant."

The limiting reagent is the reactant that is entirely provided up in a reaction, and also thus determines when the reactivity stops. From the reactivity stoichiomeattempt, the exact amount of reactant essential to react via another facet can be calculated. If the reactants are not mixed in the correct stoichiometric proparts (as shown by the balanced jiyuushikan.orgical equation), then among the reactants will be entirely consumed while another will be left over. The limiting reagent is the one that is entirely consumed; it boundaries the reaction from proceeding because tright here is none left to react with the in-excess reactant.

Tright here are 2 methods to identify the limiting reagent. One approach is to discover and also compare the mole proportion of the reactants offered in the reactivity (technique 1). Another method is to calculate the grams of products developed from the provided 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 certainly create.

Balance the jiyuushikan.orgical equation for the jiyuushikan.orgical reactivity. Convert the provided indevelopment right into moles. Use stoichiometry for each individual reactant to uncover 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 uncover the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the complete mass of excess reagent offered.

b. If every one of the 0.1388 moles of glucose were supplied up, tright here would certainly must be 0.1388 x 6 or 0.8328 moles of oxygen. Since there is an excess of oxygen, the glucose amount is offered to calculate the amount of the products in the reactivity.
Because of this, the mole proportion is: (0.8328 mol O2)/(0.208 mol C6H12O6)

This offers a 4.004 ratio of O2 to C6H12O6.

Tip 4: Use the amount of limiting reactant to calculate the amount of CO2 or H2O developed.

For carbon dioxide produced: (mathrm0.1388: moles: glucose imes dfrac61 = 0.8328: moles: carbon: dioxide).

Tip 5: If important, calculate how a lot is left in excess.

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

Tip 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 making use of 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 offered – 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 every one of the oxygen is provided up, (mathrm1.53 imes dfrac411) or 0.556 moles of C2H3Br3 are compelled. Since tright here are just 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)

Because of this, by either approach, 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 every one of the water is consumed, (mathrm1.633 imes dfrac22) or 1.633 moles of Na2O2 are required. Due to the fact that there are just 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 entirely 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. Tright here must be 1 mole of SiO2 for eincredibly 2 moles of H2F2 consumed. Due to the fact that the proportion is 0.478 to 0.568, 28.7 grams of SiO2 do not react via the H2F2.