# 4.4: Determining the Limiting Reactant

Learning Objectives

• To understand the concept of limiting reactants and quantify incomplete reactions

In all the examples discussed therefore far, the reactants were assumed to be give in stoichiometric quantities. consequently, none of the reactants was left over at the end of the reaction. This is much desirable, as in the case of a outer space shuttle, where excess oxygen or hydrogen was not entirely extra freight to be hauled into sphere but besides an explosion luck. More much, however, reactants are present in breakwater ratios that are not the same as the proportion of the coefficients in the balance chemical equation. As a consequence, one or more of them will not be used up completely but will be left over when the reaction is completed. In this situation, the total of merchandise that can be obtained is limited by the come of only one of the reactants. The reactant that restricts the measure of product obtained is called the limit reactant. The reactant that remains after a reaction has gone to completion is in excess .
Consider a nonchemical exemplar. Assume you have invited some friends for dinner and want to bake brownies for dessert. You find two boxes of elf mix in your pantry and see that each package requires two eggs. The balance equality for brownie training is thus
\ [ 1 \, \text { box mix } + 2 \, \text { eggs } \rightarrow 1 \, \text { batch brownies } \label { 4.4.1 } \ ]

If you have a twelve eggs, which ingredient will determine the number of batches of brownies that you can prepare ? Because each box of brownie desegregate requires two eggs and you have two boxes, you need four eggs. Twelve egg is eight more eggs than you need. Although the proportion of eggs to boxes in is 2:1, the proportion in your possession is 6:1. Hence the eggs are the ingredient ( reactant ) deliver in surfeit, and the brownie desegregate is the limiting reactant. evening if you had a refrigerator full of eggs, you could make only two batches of brownies . Figure $$\PageIndex{1}$$: The Concept of a Limiting Reactant in the Preparation of Brownies

now consider a chemical case of a restrict reactant : the production of saturated titanium. This alloy is fairly light ( 45 % lighter than steel and entirely 60 % heavier than aluminum ) and has great mechanical forte ( ampere hard as sword and twice deoxyadenosine monophosphate solid as aluminum ). Because it is besides highly immune to corrosion and can withstand extreme temperatures, titanium has many applications in the aerospace industry. titanium is besides used in medical implants and portable calculator housings because it is luminosity and insubordinate to corrosion. Although titanium is the ninth most park element in Earth ’ s crust, it is relatively unmanageable to extract from its ores. In the inaugural footstep of the origin process, titanium-containing oxide minerals react with solid carbon and chlorine flatulence to form titanium tetrachloride ( TiCl4 ) and carbon dioxide. Titanium tetrachloride is then converted to metallic titanium by reaction with magnesium metallic at high temperature :
\ [ TiCl_4 ( gigabyte ) + 2 \, Mg ( liter ) \rightarrow Ti ( s ) + 2 \, MgCl_2 ( fifty ) \label { 4.4.2 } \ ]
Because titanium ores, carbon, and chlorine are all quite cheap, the high price of titanium ( about \$ 100 per kilogram ) is largely ascribable to the high cost of magnesium metal. Under these circumstances, magnesium metallic is the limiting reactant in the output of metallic titanium . Figure courtesy of NIH (NIADDK) 9AO4 (Connie Raab)

With 1.00 kg of titanium tetrachloride and 200 gigabyte of magnesium metal, how much titanium alloy can be produced according to the equality above ? Solving this type of trouble requires that you carry out the come steps :

1. Determine the number of moles of each reactant.
2. Compare the mole ratio of the reactants with the ratio in the balanced chemical equation to determine which reactant is limiting.
3. Calculate the number of moles of product that can be obtained from the limiting reactant.
4. Convert the number of moles of product to mass of product.

1. To determine the number of moles of reactants give, calculate or look up their molar masses : 189.679 g/mol for titanium tetrachloride and 24.305 g/mol for magnesium. The number of moles of each is calculated as follows :
\ [ moles \, TiCl_4 = { mass \, TiCl_4 \over molar \, mass \, TiCl_4 } \ ]
\ [ = 1000 \, g \, TiCl_4 \times { 1 \, gram molecule \, TiCl_4 \over 189.679 \, g \, TiCl_4 } = 5.272 \, gram molecule \, TiCl_4 \ ]
\ [ moles \, Mg = { multitude \, Mg \over molar \, mass \, Mg } \ ]
\ [ = 200 \, gigabyte \, Mg \times { 1 \, gram molecule \, Mg \over 24.305 \, guanine \, Mg } = 8.23 \, gram molecule \, Mg \ ]
2. There are more moles of magnesium than of titanium tetrachloride, but the ratio is only the following :
\ [ { gram molecule \, Mg \over gram molecule \, TiCl_4 } = { 8.23 \, gram molecule \over 5.272 \, gram molecule } = 1.56 \ ]
Because the proportion of the coefficients in the balanced chemical equality is ,
\ [ { 2 \, gram molecule \, Mg \over 1 \, gram molecule \, TiCl_4 } = 2 \ ]
there is not have enough magnesium to react with all the titanium tetrachloride. If this point is not unclutter from the gram molecule proportion, calculate the number of moles of one reactant that is required for complete reaction of the other reactant. For example, there are 8.23 mol of Mg, sol ( 8.23 ÷ 2 ) = 4.12 gram molecule of TiCl4 are required for complete reaction. Because there are 5.272 mol of TiCl4, titanium tetrachloride is present in excess. conversely, 5.272 gram molecule of TiCl4 requires 2 × 5.272 = 10.54 gram molecule of Mg, but there are only 8.23 gram molecule. consequently, magnesium is the limiting reactant .
3. Because magnesium is the limiting reactant, the phone number of moles of magnesium determines the numeral of moles of titanium that can be formed :
\ [ moles \, Ti = 8.23 \, gram molecule \, Mg = { 1 \, gram molecule \, Ti \over 2 \, gram molecule \, Mg } = 4.12 \, gram molecule \, Ti \ ]
therefore merely 4.12 gram molecule of Ti can be formed .
4. To calculate the multitude of titanium alloy that can obtain, multiply the number of moles of titanium by the molar aggregate of titanium ( 47.867 g/mol ) :
\ [ moles \, Ti = mass \, Ti \times molar \, aggregate \, Ti = 4.12 \, gram molecule \, Ti \times { 47.867 \, gram \, Ti \over 1 \, gram molecule \, Ti } = 197 \, gravitational constant \, Ti \ ]
here is a childlike and reliable way to identify the restrict reactant in any problem of this screen :

1. Calculate the number of moles of each reactant present: 5.272 mol of TiCl4 and 8.23 mol of Mg.
2. Divide the actual number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation: $TiCl_4 : { 5.272 \, mol \, (actual) \over 1 \, mol \, (stoich)} = 5.272 \, \, \, \, Mg: {8.23 \, mol \, (actual) \over 2 \, mol \, (stoich)} = 4.12$
3. The reactant with the smallest mole ratio is limiting. Magnesium, with a calculated stoichiometric mole ratio of 4.12, is the limiting reactant.

density is the mass per unit volume of a means. If we are given the concentration of a message, we can use it in stoichiometric calculations involving fluent reactants and/or products, as Example \ ( \PageIndex { 1 } \ ) demonstrates .

Determining the Limiting Reactant and Theoretical Yield for a reaction : hypertext transfer protocol : //youtu.be/HmDm1qpNUD0
exemplar \ ( \PageIndex { 1 } \ ) : Fingernail Polish Remover
Ethyl acetate rayon ( CH3CO2C2H5 ) is the solvent in many fingernail polish removers and is used to decaffeinate coffee beans and tea leaves. It is prepared by reacting ethyl alcohol ( C2H5OH ) with acetic acid ( CH3CO2H ) ; the other product is water system. A minor total of sulphuric acid is used to accelerate the reaction, but the sulphuric acid is not consumed and does not appear in the balance chemical equation. Given 10.0 milliliter each of acetic acid and ethyl alcohol, how many grams of ethyl acetate rayon can be prepared from this reaction ? The densities of acetic acerb and ethyl alcohol are 1.0492 g/mL and 0.7893 g/mL, respectively . Given : reactants, products, and volumes and densities of reactants
Asked for : mass of product
Strategy :

1. Balance the chemical equation for the reaction.
2. Use the given densities to convert from volume to mass. Then use each molar mass to convert from mass to moles.
3. Using mole ratios, determine which substance is the limiting reactant. After identifying the limiting reactant, use mole ratios based on the number of moles of limiting reactant to determine the number of moles of product.
4. Convert from moles of product to mass of product.

Solution :
A Always begin by writing the poise chemical equation for the reaction :
\ [ C_2H_5OH ( l ) + CH_3CO_2H ( aq ) \rightarrow CH_3CO_2C_2H_5 ( aq ) + H_2O ( lambert ) \ ]
B We need to calculate the number of moles of ethyl alcohol and acetic acid that are deliver in 10.0 milliliter of each. recall from that the concentration of a substance is the mass divided by the volume :
\ [ density = { mass \over volume } \ ]
Rearranging this expression gives mass = ( density ) ( volume ). We can replace mass by the product of the density and the volume to calculate the number of moles of each message in 10.0 milliliter ( remember, 1 milliliter = 1 cm3 ) :
\ [ moles \, C_2H_5OH = { mass \, C_2H_5OH \over molar \, mass \, C_2H_5OH } \ ]
\ [ = { volume \, C_2H_5OH \times density \, C_2H_5OH \over molar \, mass \, C_2H_5OH } \ ]
\ [ = 10.0 \, milliliter \, C_2H_5OH \times { 0.7893 \, gigabyte \, C_2H_5OH \over 1 \, milliliter \, C_2H_5OH } \times { 1 \, gram molecule \, C_2H_5OH \over 46.07 \, g\, C_2H_5OH } \ ]
\ [ = 0.171 \, gram molecule \, C_2H_5OH \ ]
\ [ moles \, CH_3CO_2H = { mass \, CH_3CO_2H \over molar \, mass \, CH_3CO_2H } \ ]
\ [ = { volume \, CH_3CO_2H \times concentration \, CH_3CO_2H \over molar \, batch \, CH_3CO_2H } \ ]
\ [ = 10.0 \, milliliter \, CH_3CO_2H \times { 1.0492 \, g \, CH_3CO_2H \over 1 \, milliliter \, CH_3CO_2H } \times { 1 \, gram molecule \, CH_3CO_2H \over 60.05 \, g \, CH_3CO_2H } \ ]
\ [ = 0.175 \, gram molecule \, CH_3CO_2H \ ]
C The total of moles of acetic acid exceeds the number of moles of ethyl alcohol. Because the reactants both have coefficients of 1 in the poise chemical equation, the mole proportion is 1:1. We have 0.171 gram molecule of ethyl alcohol and 0.175 gram molecule of acetic acidic, so ethyl alcohol is the limiting reactant and acetic acerb is in excess. The coefficient in the balanced chemical equality for the product ( ethyl acetate rayon ) is besides 1, so the breakwater ratio of ethyl alcohol and ethyl acetate is besides 1:1. This means that given 0.171 gram molecule of ethyl alcohol, the measure of ethyl acetate rayon produced must besides be 0.171 mol :
\ [ moles \, ethyl \, acetate rayon = molethanol \times { 1 \, gram molecule \, ethyl \, acetate \over 1 \, gram molecule \, ethyl alcohol } \ ]
\ [ = 0.171 \, gram molecule \, C_2H_5OH \times { 1 \, gram molecule \, CH_3CO_2C_2H_5 \over 1 \, gram molecule \, C_2H_5OH } \ ]
\ [ = 0.171 \, gram molecule \, CH_3CO_2C_2H_5 \ ]
D The final step is to determine the mass of ethyl acetate rayon that can be formed, which we do by multiplying the number of moles by the molar mass :
\ [ mass \, of \, ethyl \, acetate rayon = moleethyl \, acetate rayon \times molar \, mass \, ethyl \, acetate\ ]
\ [ = 0.171 \, gram molecule \, CH_3CO_2C_2H_5 \times { 88.11 \, thousand \, CH_3CO_2C_2H_5 \over 1 \, gram molecule \, CH_3CO_2C_2H_5 } \ ]
\ [ = 15.1 \, gigabyte \, CH_3CO_2C_2H_5 \ ]
thus 15.1 thousand of ethyl acetate can be prepared in this reaction. If necessary, you could use the concentration of ethyl acetate rayon ( 0.9003 g/cm3 ) to determine the volume of ethyl acetate that could be produced :
\ [ bulk \, of \, ethyl \, acetate = 15.1 \, gravitational constant \, CH_3CO_2C_2H_5 \times { 1 \, milliliter \, CH_3CO_2C_2H_5 \over 0.9003 \, g\, CH_3CO_2C_2H_5 } \ ]
\ [ = 16.8 \, milliliter \, CH_3CO_2C_2H_5 \ ]
exercise \ ( \PageIndex { 1 } \ )
Under appropriate conditions, the reaction of elemental morning star and elemental sulfur produces the compound P4S10. How a lot P4S10 can be prepared starting with 10.0 g of P4 and 30.0 g of S8 ?

## Limiting Reactants in Solutions

The concept of limiting reactants applies to reactions carried out in solution american samoa well as to reactions involving pure substances. If all the reactants but one are present in excess, then the amount of the limiting reactant may be calculated as illustrated in Example \ ( \PageIndex { 2 } \ ) .
example
Because the pulmonary tuberculosis of alcoholic beverages adversely affects the performance of tasks that require skill and sagacity, in most countries it is illegal to drive while under the charm of alcohol. In about all US states, a blood alcohol level of 0.08 % by bulk is considered legally drunk. Higher levels cause acute accent poisoning ( 0.20 % ), unconsciousness ( about 0.30 % ), and evening death ( about 0.50 % ). The Breathalyzer is a portable device that measures the ethyl alcohol concentration in a person ’ sulfur breath, which is directly proportional to the blood alcohol degree. The reaction used in the Breathalyzer is the oxidation of ethyl alcohol by the bichromate ion :
\ [ 3CH_3 CH_2 OH ( aq ) + \underset { yellow-orange } { 2Cr_2 O_7^ { 2 – } } ( aq ) + 16H ^+ ( aq ) \underset { H_2 SO_4 ( aq ) } { \xrightarrow { \hspace { 10px } Ag ^+\hspace { 10px } } } 3CH_3 CO_2 H ( aq ) + \underset { greens } { 4Cr^ { 3+ } } ( aq ) + 11H_2 O ( fifty ) \ ]
When a measured volume ( 52.5 milliliter ) of a defendant ’ randomness hint is bubbled through a solution of excess potassium bichromate in diluted sulphuric acid, the ethyl alcohol is quickly absorb and oxidized to acetic acid by the bichromate ions. In the process, the chromium atoms in some of the Cr2O72− ions are reduced from Cr6+ to Cr3+. In the presence of Ag+ ions that act as a catalyst, the reaction is complete in less than a moment. Because the Cr2O72− ion ( the reactant ) is yellow-orange and the Cr3+ ion ( the product ) forms a green solution, the amount of ethyl alcohol in the person ’ s breath ( the confining reactant ) can be determined quite accurately by comparing the color of the final solution with the colors of standard solutions prepared with know amounts of ethyl alcohol . A Breathalyzer reaction with a test tube before (a) and after (b) ethanol is added. When a measured volume of a distrust ’ sulfur hint is bubbled through the solution, the ethyl alcohol is oxidized to acetic acid, and the solution changes color from yellow-orange to green. The intensity of the fleeceable color indicates the sum of ethyl alcohol in the sample .
A typical Breathalyzer phial contains 3.0 milliliter of a 0.25 mg/mL solution of K2Cr2O7 in 50 % H2SO4 ampere well as a cook concentration of AgNO3 ( typically 0.25 mg/mL is used for this purpose ). How many grams of ethyl alcohol must be show in 52.5 milliliter of a person ’ s breath to convert all the Cr6+ to Cr3+ ?
Given: volume and concentration of one reactant
Asked for: mass of other reactant needed for complete reaction
Strategy:

1. Calculate the number of moles of Cr2O72− ion in 1 mL of the Breathalyzer solution by dividing the mass of K2Cr2O7 by its molar mass.
2. Find the total number of moles of Cr2O72− ion in the Breathalyzer ampul by multiplying the number of moles contained in 1 mL by the total volume of the Breathalyzer solution (3.0 mL).
3. Use the mole ratios from the balanced chemical equation to calculate the number of moles of C2H5OH needed to react completely with the number of moles of Cr2O72− ions present. Then find the mass of C2H5OH needed by multiplying the number of moles of C2H5OH by its molar mass.

Solution:
A In any stoichiometry trouble, the first step is always to calculate the issue of moles of each reactant give. In this case, we are given the mass of K2Cr2O7 in 1 milliliter of solution, which can be used to calculate the act of moles of K2Cr2O7 contained in 1 mL :
\ ( \dfrac { moles\ : K_2 Cr_2 O_7 } { 1\ : milliliter } = \dfrac { ( 0 .25\ : \cancel { milligram } \ : K_2 Cr_2 O_7 ) } { milliliter } \left ( \dfrac { 1\ : \cancel { deoxyguanosine monophosphate } } { 1000\ : \cancel { milligram } } \right ) \left ( \dfrac { 1\ : gram molecule } { 294 .18\ : \cancel { gigabyte } \ : K_2 Cr_2 O_7 } \right ) = 8.5 \times 10 ^ { -7 } \ : moles \ )
B Because 1 gram molecule of K2Cr2O7 produces 1 gram molecule of Cr2O72− when it dissolves, each milliliter of solution contains 8.5 × 10−7 gram molecule of Cr2O72−. The total number of moles of Cr2O72− in a 3.0 milliliter Breathalyzer phial is frankincense
\ ( moles\ : Cr_2 O_7^ { 2- } = \left ( \dfrac { 8 .5 \times 10^ { -7 } \ : gram molecule } { 1\ : \cancel { milliliter } } \right ) ( 3 .0\ : \cancel { mL } ) = 2 .6 \times 10^ { -6 } \ : mol\ : Cr_2 O_7^ { 2– } \ )
C The balance chemical equation tells us that 3 gram molecule of C2H5OH is needed to consume 2 gram molecule of Cr2O72− ion, so the sum number of moles of C2H5OH required for dispatch reaction is
\ ( moles\ : of\ : C_2 H_5 OH = ( 2.6 \times 10 ^ { -6 } \ : \cancel { mol\ : Cr_2 O_7 ^ { 2- } } ) \left ( \dfrac { 3\ : mol\ : C_2 H_5 OH } { 2\ : \cancel { mol\ : Cr _2 O _7 ^ { 2 – } } } \right ) = 3 .9 \times 10 ^ { -6 } \ : mol\ : C _2 H _5 OH \ )
As indicated in the strategy, this number can be converted to the mass of C2H5OH using its molar mass :
\ ( mass\ : C _2 H _5 OH = ( 3 .9 \times 10 ^ { -6 } \ : \cancel { mol\ : C _2 H _5 OH } ) \left ( \dfrac { 46 .07\ : guanine } { \cancel { mol\ : C _2 H _5 OH } } \right ) = 1 .8 \times 10 ^ { -4 } \ : g\ : C _2 H _5 OH \ )
therefore 1.8 × 10−4 g or 0.18 milligram of C2H5OH must be present. experimentally, it is found that this value corresponds to a blood alcohol level of 0.7 %, which is normally black .
exercise \ ( \PageIndex { 2 } \ )
The compound para-nitrophenol ( molar mass = 139 g/mol ) reacts with sodium hydroxide in aqueous solution to generate a yellow anion via the reaction Because the sum of para-nitrophenol is easily estimated from the intensity of the yellow discolor that results when excess NaOH is added, reactions that produce para-nitrophenol are normally used to measure the natural process of enzymes, the catalysts in biological systems. What bulk of 0.105 M NaOH must be added to 50.0 milliliter of a solution containing 7.20 × 10−4 guanine of para-nitrophenol to ensure that formation of the yellow anion is complete ?
Answer: 4.93 × 10−5 L or 49.3 μL
In Examples 4.4.1 and 4.4.2, the identities of the confining reactants are apparent : [ Au ( CN ) 2 ] −, LaCl3, ethyl alcohol, and para-nitrophenol. When the limiting reactant is not apparent, it can be determined by comparing the molar amounts of the reactants with their coefficients in the balance chemical equation. The lone difference is that the volumes and concentrations of solutions of reactants, rather than the masses of reactants, are used to calculate the number of moles of reactants, as illustrated in Example \ ( \PageIndex { 3 } \ ) .
exercise \ ( \PageIndex { 3 } \ )
When aqueous solutions of eloquent nitrate and potassium bichromate are shuffle, an rally reaction occurs, and flatware bichromate is obtained as a crimson solid. The overall chemical equation for the chemical reaction is as follows :
\ ( 2AgNO_3 ( aq ) + K_2Cr_2O_7 ( aq ) \rightarrow Ag_2Cr_2O_7 ( s ) + 2KNO_3 ( aq ) \ )
What batch of Ag2Cr2O7 is formed when 500 milliliter of 0.17 M K2Cr2O7 are desegregate with 250 mL of 0.57 M AgNO3 ?
Given: balanced chemical equality and volume and concentration of each reactant
Strategy:

1. Calculate the number of moles of each reactant by multiplying the volume of each solution by its molarity.
2. Determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation.
3. Use mole ratios to calculate the number of moles of product that can be formed from the limiting reactant. Multiply the number of moles of the product by its molar mass to obtain the corresponding mass of product.

Solution:
A The balanced chemical equation tells us that 2 gram molecule of AgNO3 ( aq ) reacts with 1 mol of K2Cr2O7 ( aq ) to form 1 gram molecule of Ag2Cr2O7 ( sulfur ) ( Figure \ ( \PageIndex { 2 } \ ) ). The first step is to calculate the number of moles of each reactant in the assign volumes :
\ [ moles\ : K_2 Cr_2 O_7 = 500\ : \cancel { milliliter } \left ( \dfrac { 1\ : \cancel { L } } { 1000\ : \cancel { milliliter } } \right ) \left ( \dfrac { 0 .17\ : mol\ : K_2 Cr_2 O_7 } { 1\ : \cancel { L } } \right ) = 0 .085\ : mol\ : K_2 Cr_2 O_7 \ ]
\ [ moles\ : AgNO_3 = 250\ : \cancel { milliliter } \left ( \dfrac { 1\ : \cancel { L } } { 1000\ : \cancel { milliliter } } \right ) \left ( \dfrac { 0 .57\ : mol\ : AgNO_3 } { 1\ : \cancel { L } } \right ) = 0 .14\ : mol\ : AgNO_3 \ ]
B immediately determine which reactant is limiting by dividing the count of moles of each reactant by its stoichiometric coefficient :
\ [ K_2 Cr_2 O_7 : \ : \dfrac { 0 .085\ : gram molecule } { 1\ : gram molecule } = 0 .085 \ ]
\ [ AgNO_3 : \ : \dfrac { 0 .14\ : gram molecule } { 2\ : gram molecule } = 0 .070 \ ]
Because 0.070 < 0.085, we know that AgNO3 is the limiting reactant . C Each mole of Ag2Cr2O7 formed requires 2 gram molecule of the restrict reactant ( AgNO3 ), so we can obtain lone 0.14/2 = 0.070 gram molecule of Ag2Cr2O7. finally, convert the number of moles of Ag2Cr2O7 to the correspond batch :
\ [ mass\ : of\ : Ag_2 Cr_2 O_7 = 0 .070\ : \cancel { gram molecule } \left ( \dfrac { 431 .72\ : gravitational constant } { 1 \ : \cancel { gram molecule } } \right ) = 30\ : gigabyte \ : Ag_2 Cr_2 O_7 \ ]
The Ag+ and Cr2O72− ions form a red hasty of solid Ag2Cr2O7, while the K+ and NO3− ions remain in solution. ( water molecules are omitted from molecular views of the solutions for clarity. )
exercise \ ( \PageIndex { 3 } \ )
aqueous solutions of sodium bicarbonate and sulphuric acerb react to produce carbon paper dioxide according to the following equality :
\ ( 2NaHCO_3 ( aq ) + H_2SO_4 ( aq ) \rightarrow 2CO_2 ( deoxyguanosine monophosphate ) + Na_2SO_4 ( aq ) + 2H_2O ( lambert ) \ )
If 13.0 milliliter of 3.0 M H2SO4 are added to 732 milliliter of 0.112 M NaHCO3, what bulk of CO2 is produced ?