How To Find The Limiting Reactant With Moles

Limiting Reagents

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When there is not enough of one reactant in a chemical reaction, the reaction stops abruptly. To effigy out the corporeality of product produced, it must be determined reactant will limit the chemic reaction (the limiting reagent) and which reactant is in excess (the excess reagent). One style of finding the limiting reagent is by calculating the corporeality of product that can be formed past each reactant; the one that produces less production is the limiting reagent.

Introduction

The post-obit scenario illustrates the significance of limiting reagents. In society to gather a car, iv tires and 2 headlights are needed (among other things). In this example, imagine that the tires and headlights are reactants while the auto is the production formed from the reaction of 4 tires and 2 headlights. If yous have 20 tires and fourteen headlights, how many cars tin be made? With 20 tires, five cars can be produced because at that place are 4 tires to a car. With 14 headlights, vii cars can exist built (each car needs 2 headlights). Although more cars can be made from the headlights available, simply 5 full cars are possible considering of the limited number of tires bachelor. In this example, the headlights are in excess. Considering the number of cars formed past 20 tires is less than number of cars produced past 14 headlights, the tires are the limiting reagent (they limit the full completion of the reaction, in which all of the reactants are used upward). This scenario is illustrated below:

iv Tires + 2 Headlights = 1 Car

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

800px-Jaguar_E-type_(serie_III)_-_headlamp.jpg — Figure 1: The synthesis reaction of making a car. Images used from Wikipedia with permission.

The initial condition is that there must exist 4 tires to 2 headlights. The reactants must thus occur in that ratio; otherwise, one will limit the reaction. There are twenty tires and 14 headlights, and so there are ii ways of looking at this problem. For 20 tires, x headlights are required, whereas for 14 headlights, 28 tires are required. Because there are non 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 upwards in a reaction, and thus determines when the reaction stops. From the reaction stoichiometry, the verbal corporeality of reactant needed to react with another element can exist calculated. If the reactants are not mixed in the correct stoichiometric proportions (as indicated by the balanced chemical equation), then one of the reactants volition exist entirely consumed while another will be left over. The limiting reagent is the one that is totally consumed; information technology limits the reaction from standing considering there is none left to react with the in-excess reactant.

There are two ways to decide the limiting reagent. I 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 corporeality of production is the limiting reagent (approach 2).

How to Detect the Limiting Reagent: Arroyo i

Detect the limiting reagent by looking at the number of moles of each reactant.

Determine the balanced chemical equation for the chemical reaction.
Catechumen all given data into moles (most likely, through the use of molar mass equally a conversion factor).
Summate the mole ratio from the given data. Compare the calculated ratio to the actual ratio.
Utilise the amount of limiting reactant to calculate the corporeality of product produced.
If necessary, calculate how much is left in excess of the not-limiting reagent.

How to Find the Limiting Reagent: Approach 2

Discover the limiting reagent by computing and comparing the corporeality of product each reactant volition produce.

Residue the chemical equation for the chemical reaction.
Convert the given data into moles.
Use stoichiometry for each individual reactant to find the mass of production produced.
The reactant that produces a bottom amount of product is the limiting reagent.
The reactant that produces a larger amount of product is the backlog reagent.
To find the amount of remaining excess reactant, subtract the mass of backlog reagent consumed from the total mass of excess reagent given.

Case \(\PageIndex{1}\): Photosynthesis

Consider respiration, one of the well-nigh mutual chemic reactions on earth.

\[\ce{ C6H_{12}O6 + 6 O_2 \rightarrow half dozen CO2 + six H2o} + \rm{energy}\]

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

Solution

When approaching this problem, find that every ane mole of glucose (\(C_6H_{12}O_6\)) requires 6 moles of oxygen to obtain 6 moles of carbon dioxide and 6 moles of water.

Step one: Decide the balanced chemic equation for the chemical reaction.

The balanced chemic equation is already given.

Step ii: Convert all given information into moles (well-nigh likely, through the use of molar mass as a conversion factor).

\(\mathrm{25\:g \times \dfrac{1\: mol}{180.06\:g} = 0.1388\: mol\: C_6H_{12}O_6}\)

\(\mathrm{forty\:g \times \dfrac{1\: mol}{32\:g} = 1.25\: mol\: O_2}\)

Step three: Summate the mole ratio from the given information. Compare the calculated ratio to the bodily ratio.

a. If all of the 1.25 moles of oxygen were to be used up, in that location would need to be \(\mathrm{ane.25 \times \dfrac{ane}{six}}\) or 0.208 moles of glucose. At that place is but 0.1388 moles of glucose available which makes it the limiting reactant.

\[ane.25 \; \rm{mol} \; O_2 \times \dfrac{ 1 \; \rm{mol} \; C_6H_{12}O_6}{6\; \rm{mol} \; O_2}= 0.208 \; \rm{mol} \; C_6H_{12}O_6 \nonumber\]

b. If all of the 0.1388 moles of glucose were used up, in that location would demand to exist 0.1388 x 6 or 0.8328 moles of oxygen. Because there is an backlog of oxygen, the glucose amount is used to summate the amount of the products in the reaction.

\[0.1388\; \rm{ mol}\; C_6H_{12}O_6 \times \dfrac{vi \; \rm{mol} \;O_2}{ane \; \rm{mol} \; C_6H_{12}O_6} = 0.8328 \; \rm{mol}\; O_2 \nonumber\]

If more than than half dozen moles of O_iiare available per mole of C₆H₁₂O_half-dozen, the oxygen is in excess and glucose is the limiting reactant. If less than 6 moles of oxygen are available per mole of glucose, oxygen is the limiting reactant. The ratio is 6 mole oxygen per one mole glucose, OR 1 mole oxygen per one/6 mole glucose. This ways: 6 mol O₂/ 1 mol C₆H₁₂O₆ .

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

This gives a four.004 ratio of O _{2 to} C₆H₁₂O₆ _.

Pace 4: Use the amount of limiting reactant to summate the amount of CO₂ or H_iiO produced.

For carbon dioxide produced: \(\mathrm{0.1388\: moles\: glucose \times \dfrac{6}{one} = 0.8328\: moles\: carbon\: dioxide}\).

Pace 5: If necessary, summate how much is left in backlog.

ane.25 mol - 0.8328 mol = 0.4172 moles of oxygen left over

Case \(\PageIndex{2}\): Oxidation of Magnesium

Calculate the mass of magnesium oxide possible if 2.40 g \(Mg \) reacts with 10.0 g \(O_2\)

\[\ce{ Mg +O_2 \rightarrow MgO} \nonumber\]

Solution

Step 1: Remainder equation

\[\ce{two Mg + O_2 \rightarrow two MgO} \nonumber\]

Stride ii and Step iii: Converting mass to moles and stoichiometry

\(\mathrm{2.40\:chiliad\: Mg \times \dfrac{1.00\: mol\: Mg}{24.31\:g\: Mg} \times \dfrac{2.00\: mol\: MgO}{2.00\: mol\: Mg} \times \dfrac{40.31\:g\: MgO}{1.00\: mol\: MgO} = 3.98\:m\: MgO}\)

\(\mathrm{10.0\:thousand\: O_2\times \dfrac{1\: mol\: O_2}{32.0\:g\: O_2} \times \dfrac{two\: mol\: MgO}{1\: mol\: O_2} \times \dfrac{40.31\:one thousand\: MgO}{ane\: mol\: MgO} = 25.2\: g\: MgO}\)

Step 4: The reactant that produces a smaller amount of production is the limiting reagent

Mg produces less MgO than does O₂(3.98 grand MgO vs. 25.2 thou MgO), therefore Mg is the limiting reagent in this reaction.

Step five: The reactant that produces a larger amount of production is the backlog reagent

O₂ produces more amount of MgO than Mg (25.2g MgO vs. iii.98 MgO), therefore O₂ is the excess reagent in this reaction.

Pace six: 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 backlog reagent calculated using the limiting reagent:

\(\mathrm{2.40\:yard\: Mg \times \dfrac{i.00\: mol\: Mg}{24.31\:grand\: Mg} \times \dfrac{i.00\: mol\: O_2}{two.00\: mol\: Mg} \times \dfrac{32.0\:m\: O_2}{1.00\: mol\: O_2} = 1.58\:g\: O_2}\)
OR Mass of excess reagent calculated using the mass of the product:
\(\mathrm{3.98\:g\: MgO \times \dfrac{i.00\: mol\: MgO}{40.31\:thousand\: MgO} \times \dfrac{1.00\: mol\: O_2}{2.00\: mol\: MgO} \times \dfrac{32.0\:g\: O_2}{1.00\: mol\: O_2} = 1.58\:thou\: O_2}\) Mass of total excess reagent given – mass of excess reagent consumed in the reaction
10.0g – 1.58g = viii.42g O₂ is in excess.

Case \(\PageIndex{three}\): Limiting Reagent

What is the limiting reagent if 76.four grams of \(C_2H_3Br_3\) were reacted with 49.1 grams of \(O_2\)?

\[\ce{4 C_2H_3Br_3 + xi O_2 \rightarrow 8 CO_2 + 6 H_2O + 6 Br_2} \nonumber\]

Solution

Using Approach 1:

A. \(\mathrm{76.4\:g \times \dfrac{1\: mole}{266.72\:m} = 0.286\: moles\: of\: C_2H_3Br_3}\)

\(\mathrm{49.ane\: g \times \dfrac{i\: mole}{32\:g} = i.53\: moles\: of\: O_2}\)

B. Assuming that all of the oxygen is used up, \(\mathrm{1.53 \times \dfrac{four}{11}}\) or 0.556 moles of C_iiH_iiiBr_iii are required. Because in that location are only 0.286 moles of C₂H_threeBr_iii available, C_twoH_threeBr_three is the limiting reagent.

Using Approach ii:

\(\mathrm{76.4\:g\: C_2H_3Br_3 \times \dfrac{1\: mol\: C_2H_3Br_3}{266.72\:1000\: C_2H_3Br_3} \times \dfrac{8\: mol\: CO_2}{4\: mol\: C_2H_3Br_3} \times \dfrac{44.01\:one thousand\: CO_2}{1\: mol\: CO_2} = 25.2\:g\: CO_2}\)

\(\mathrm{49.one\:1000\: O_2 \times \dfrac{1\: mol\: O_2}{32\:thousand\: O_2} \times \dfrac{8\: mol\: CO_2}{11\: mol\: O_2} \times \dfrac{44.01\:g\: CO_2}{1\: mol\: CO_2} = 49.1\:1000\: CO_2}\)

Therefore, past either method, C₂H₃Br₃is the limiting reagent.

Example \(\PageIndex{iv}\): Limiting Reagent

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

Solution

Using Approach 1:

A. \(\mathrm{78\:g \times \dfrac{1\: mol}{77.96\:g} = 1.001\: moles\: of\: Na_2O_2}\)

\(\mathrm{29.4\:g \times \dfrac{1\: mol}{18\:yard}= 1.633\: moles\: of\: H_2O}\)

B. Presume that all of the water is consumed, \(\mathrm{1.633 \times \dfrac{2}{2}}\) 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:

\(\mathrm{78\:g\: Na_2O_2 \times \dfrac{1\: mol\: Na_2O_2}{77.96\:g\: Na_2O_2} \times \dfrac{4\: mol\: NaOH}{2\: mol\: Na_2O_2} \times \dfrac{xl\:one thousand\: NaOH}{ane\: mol\: NaOH} = 80.04\:one thousand\: NaOH}\)

Using either approach gives Na₂O_ii as the limiting reagent.

Example \(\PageIndex{5}\): Excess Reagent

How much the backlog reagent remains if 24.v grams of CoO is reacted with 2.58 grams of O₂?

\[iv CoO + O_2 \rightarrow 2 Co_2O_3\]

Solution

A. \(\mathrm{24.v\:g \times \dfrac{1\: mole}{74.9\:g}= 0.327\: moles\: of\: CoO}\)

\(\mathrm{2.58\:m \times \dfrac{1\: mole}{32\:m}= 0.0806\: moles\: of\: O_2}\)

B. Assuming that all of the oxygen is used up, \(\mathrm{0.0806 \times \dfrac{iv}{1}}\) or 0.3225 moles of \(CoO\) are required. Because 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.

Instance \(\PageIndex{half dozen}\): Identifying the Limiting Reagent

Will 28.7 grams of \(SiO_2\) react completely with 22.half dozen grams of \(H_2F_2\)? If not, identify the limiting reagent.

\[SiO_2+ two H_2F_2 \rightarrow SiF_4+ 2 H_2O\]

Solution

A. \(\mathrm{28.7\:g \times \dfrac{1\: mole}{60.08\:g} = 0.478\: moles\: of\: SiO_2}\)

\(\mathrm{22.6\:g \times \dfrac{one\: mole}{39.eight\:g} = 0.568\: moles\: of\: H_2F_2}\)

B. At that place must be 1 mole of SiO₂ for every ii moles of H_iiF₂ consumed. Because the ratio is 0.478 to 0.568, 28.7 grams of SiO₂ do not react with the H₂F₂.

C. Assuming that all of the silicon dioxide is used up, \(\mathrm{0.478 \times \dfrac{2}{one}}\) or 0.956 moles of H_twoF₂ are required. Because there are only 0.568 moles of H₂F_two, information technology is the limiting reagent .

References

Petrucci, Ralph H., William Southward. Harwood, Geoffery F. Herring, and Jeffry D. Madura. General Chemistry. 9th ed. New Bailiwick of jersey: Pearsin Prentice Hall, 2007.
Staley, Dennis. Prentice Hall Chemical science. Boston: Pearson Prentice Hall, 2007.

Contributors and Attributions

Sarick Shah (UCD)

Source: https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Limiting_Reagents

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