How To Find Limiting Reagent With Moles

Learning Objective

Decide the limiting reagent and the amount of a product formed in a given reavion

Cardinal Points

The limiting reagent is the reactant that is used upward completely. This stops the reaction and no further products are made.
Given the balanced chemical equation that describes the reaction, there are several ways to identify the limiting reagent.
Ane style to determine the limiting reagent is to compare the mole ratios of the amounts of reactants used. This method is nearly useful when at that place are merely 2 reactants.
The limiting reagent tin can also exist derived by comparison the corporeality of products that can be formed from each reactant.

Term

limiting reagentThe reactant in a chemic reaction that is consumed first; prevents any further reaction from occurring.

In a chemic reaction, the limiting reagent, or limiting reactant, is the substance that has been completely consumed when the chemical reaction is consummate. The amount of product produced by the reaction is limited by this reactant because the reaction cannot proceed further without it; oftentimes, other reagents are present in excess of the quantities required to to react with the limiting reagent. From stoichiometry, the exact amount of reactant needed to react with another element tin can be calculated. However, if the reagents are not mixed or present in these correct stoichiometric proportions, the limiting reagent will exist entirely consumed and the reaction volition non go to stoichiometric completion.

**Limiting reagent**The limiting reagent in a reaction is the first to be completely used upwards and prevents any further reaction from occurring. In this reaction, reactant B is the limiting reagent because there is still some left over A in the products. Therefore, A was in excess when B was all used up.

Determining the Limiting Reagent

One mode to determine the limiting reagent is to compare the mole ratio of the corporeality of reactants used. This method is most useful when there are only two reactants. One reactant (A) is chosen, and the balanced chemical equation is used to determine the amount of the other reactant (B) necessary to react with A. If the corporeality of B really nowadays exceeds the amount required, and so B is in excess, and A is the limiting reagent. If the amount of B present is less than is required, and then B is the limiting reagent.

To begin, the chemical equation must first be balanced. The law of conservation states that the quantity of each element does not modify over the course of a chemical reaction. Therefore, the chemic equation is balanced when the amount of each element is the aforementioned on both the left and right sides of the equation. Adjacent, catechumen all given data (typically masses) into moles, and compare the mole ratios of the given information to those in the chemical equation.

For example: What would exist the limiting reagent if 75 grams of C₂H₃Br₃ reacted with 50.0 grams of O₂ in the following reaction:

[latex]4 \ C_2H_3Br_3 + eleven \ O_2 \rightarrow 8 \ CO_2 + half dozen \ H_2O + 6 \ Br_2[/latex]

Showtime, convert the values to moles:

[latex]75 \ g \times \frac{ane \ mole}{266.72 \ g} = 0.28 \ mol \ C_2H_3Br_3[/latex]

[latex]50.0 \ g \times \frac{1 \ mol}{32 \ yard} = 1.56 \ mol \ O_2[/latex]

Information technology is then possible to calculate how much C₂H₃Br₃ would exist required if all the O₂ is used upward:

[latex]ane.56 \ mol \ O_2 \times \frac{4 \ mol \ C_2H_3Br_3}{11 \ mol \ O_2} = 0.567 \ mol \ C_2H_3Br_3[/latex]

This demonstrates that 0.567 mol C₂H₃Br₃is required to react with all the oxygen. Since at that place is only 0.28 mol C₂H_threeBr₃ present, C₂H₃Br_three is the limiting reagent.

Another method of determining the limiting reagent involves the comparing of production amounts that tin be formed from each reactant. This method can exist extended to any number of reactants more hands than the previous method. Again, begin by balancing the chemical equation and past converting all the given information into moles. Then utilise stoichiometry to calculate the mass of the product that could exist produced for each individual reactant. The reactant that produces the least amount of product is the limiting reagent.

For example: What would exist the limiting reagent if 80.0 grams of Na₂O₂ reacted with 30.0 grams of H₂O in the reaction?

[latex]2 \ Na_2O_2 + 2 \ H_2O \rightarrow 4 \ NaOH + O_2[/latex]

The comparison can be done with either product; for this example, NaOH will be the production compared. To make up one's mind how much NaOH is produced past each reagent, utilize the stoichiometric ratio given in the chemical equation as a conversion gene:

[latex]\frac{four \ mol \ NaOH}{two \ mol \ Na_2O_2}[/latex] and [latex]\frac {4 \ mol \ NaOH}{ii \ mol \ H_2O}[/latex]

And so catechumen the grams of each reactant into moles of NaOH to see how much NaOH each could produce if the other reactant was in excess.

[latex]eighty.0 \ g \ Na_2O_2 \times \frac {1 \ mol \ Na_2O_2}{77.98 \ grand \ Na_2O_2} \times \frac {four \ moles \ NaOH}{two \ mol \ Na_2O_2} = 2.06 \ moles \ NaOH[/latex]

[latex]30.0 \ 1000 \ H_2O \times \frac {1 \ mol \ H_2O}{18 \ thou \ H_2O} \times \frac {four \ moles \ NaOH}{2 \ moles \ H_2O} = three.33 \ moles \ NaOH [/latex]

Manifestly the Na_twoO₂ produces less NaOH than H₂O; therefore, Na_iiO_two is the limiting reagent.

STOICHIOMETRY – Limiting Reactant & Backlog Reactant Stoichiometry & Moles – YouTubeA video showing two examples of how to solve limiting reactant stoichiometry bug. This video also explains how to determine the excess reactant also.