StoichiometryStoichiometry

StoichiometryStoichiometry is the branch of is the branch of chemistry that deals with the quantities chemistry that deals with the quantities of substances that enter into, and are of substances that enter into, and are produced by, chemical reactions. produced by, chemical reactions. 

For example, when methane unites For example, when methane unites with oxygen in complete combustion, with oxygen in complete combustion, 16g of methane require 64g of 16g of methane require 64g of oxygen.  At the same time 44g of oxygen.  At the same time 44g of carbon dioxide and 36g of water are carbon dioxide and 36g of water are formed as reaction products. formed as reaction products.

Hamburger AnalogyHamburger Analogy My recipe for a bacon double cheeseburger is:My recipe for a bacon double cheeseburger is:

1 hamburger bun 1 hamburger bun 2 hamburger patties 2 hamburger patties 2 slices of cheese 2 slices of cheese 4 strips of bacon4 strips of bacon

Based on this recipe:Based on this recipe: If I have five bacon double cheeseburgers: If I have five bacon double cheeseburgers:

– How many hamburger buns do I have? How many hamburger buns do I have? – How many hamburger patties do I have? How many hamburger patties do I have? – How many slices of cheese do I have? How many slices of cheese do I have?

How many bacon double cheeseburgers can you How many bacon double cheeseburgers can you make if you start with: make if you start with: – 2 bun, 2 patties, 2 slices of cheese, 6 strips of bacon 2 bun, 2 patties, 2 slices of cheese, 6 strips of bacon – 2 bun, 4 patties, 4 slices of cheese, 8 strips of bacon 2 bun, 4 patties, 4 slices of cheese, 8 strips of bacon

When making hamburgers, we simply When making hamburgers, we simply used the relationship between the used the relationship between the numbers of each item necessary. The numbers of each item necessary. The numbers represented our quantities.numbers represented our quantities.

When dealing with chemical reactions, When dealing with chemical reactions, we use the same basic principle but we use the same basic principle but must use balanced equations and must use balanced equations and conversion bridges to enable us to conversion bridges to enable us to convert between quantities of chemical convert between quantities of chemical substances.substances.

Chemical Reaction ReviewChemical Reaction Review You will often have to create formula You will often have to create formula

equations given only partial information.equations given only partial information. In order to obtain a balanced equation, In order to obtain a balanced equation,

which is necessary for all stoic problems, which is necessary for all stoic problems, let us review Rx types:let us review Rx types:1. Synthesis Rx: A + B 1. Synthesis Rx: A + B AB AB2. Decomposition Rx: AB 2. Decomposition Rx: AB A + B A + B3. Single Replacement: A + XY 3. Single Replacement: A + XY AY + X AY + X4. Double Replacement: AB + XY 4. Double Replacement: AB + XY AY + AY + XBXB

5. Combustion Rx: C5. Combustion Rx: CxxHHxx + O + O22 CO CO22 + H + H22OO

Identify each rx type and predict products Identify each rx type and predict products for:for:

1.1. Magnesium ribbon and oxygen gas react:Magnesium ribbon and oxygen gas react:2.2. Silver nitrate combines with iron(III) Silver nitrate combines with iron(III)

sulfate:sulfate:

3.3. The complete combustion of CThe complete combustion of C22HH66::4.4. The decomposition of sodium chlorate The decomposition of sodium chlorate

upon heating:upon heating:5.5. Sodium metal and chlorine gas react:Sodium metal and chlorine gas react:6.6. Aluminum reacts with hydrochloric acid:Aluminum reacts with hydrochloric acid:7.7. Silver acetate reacts with magnesium Silver acetate reacts with magnesium

ribbon:ribbon:8.8. The decomposition of calcium hydroxide:The decomposition of calcium hydroxide:

Conversion BridgesConversion Bridges

2C2C22HH22(g) + 5O(g) + 5O22(g) (g) 4CO 4CO22(g) + 2 H(g) + 2 H22O(l) O(l)

Molar mass: Remember, molar mass is Molar mass: Remember, molar mass is the mass of the mass of 1 mole1 mole of any substance. of any substance.

CC22HH22 = 26.04 g/mol, O = 26.04 g/mol, O22 = 32.00 g/mol = 32.00 g/mol Molar ratios: Use the coefficients of the Molar ratios: Use the coefficients of the

balanced equation to establish mole balanced equation to establish mole ratios for any of the substances.ratios for any of the substances.

CC22HH22 to CO to CO22 = 2:4 = 1:2 = 2:4 = 1:2

HH22O to OO to O22 = 2:5 = 2:5

8Al + 3Fe8Al + 3Fe33OO44 4Al 4Al22OO33 + 9 + 9 Fe Fe

Using the balanced equation:Using the balanced equation:

1.1. Calculate the molar masses of each Calculate the molar masses of each of the substances.of the substances.

2.2. Write the molar ratios for:Write the molar ratios for:Al to FeAl to Fe

FeFe33OO44 to Fe to Fe

AlAl22OO33 to Al to Al

FeFe33OO44 to Al to Al22OO33

Types of Stoic ProblemsTypes of Stoic Problems

There are several types of There are several types of stoichiometry problems based on what stoichiometry problems based on what is given and unknown in a problem.is given and unknown in a problem.

1. Mass 1. Mass Mass Mass

2. Mole 2. Mole Mass Mass

3. Mole 3. Mole Mole Mole

4. Limiting and Excess Reactants4. Limiting and Excess Reactants

5. Theoretical Yield of Products5. Theoretical Yield of Products

Mass Mass Mass Mass

Given the equation: HGiven the equation: H22(g) + O(g) + O22(g) (g) H H22O(g)O(g)

Calculate grams of HCalculate grams of H22O produced if 25 O produced if 25 grams of oxygen is reacted with excess grams of oxygen is reacted with excess hydrogen.hydrogen.

Step 1: Start with a balanced equation.Step 1: Start with a balanced equation.

Step 2: Identify given and unknown.Step 2: Identify given and unknown.

Step 3: Complete table.Step 3: Complete table.

2H2H22(g) + O(g) + O22(g) (g) 2H 2H22O(g)O(g)Given/Given/

UnknownUnknownAmountAmount MolesMoles

(based on (based on coefficients)coefficients)

Molar Molar

MassMass

OO22 25 g25 g 11 32.00 32.00 g/molg/mol

HH22OO XX 22 18.02 18.02 g/molg/mol

*There are only two substances that will be calculated in a stoic problem, the given and unknown.

*When identifying given and unknown, make sure the chemical formulas are correct. EX: H2 not H

Step 4: Place given over 1.Step 4: Place given over 1.Step 5: Convert from grams of given to Step 5: Convert from grams of given to

moles of given using molar mass.moles of given using molar mass.Step 6: Convert from moles of given to Step 6: Convert from moles of given to

moles of unknown using molar ratio moles of unknown using molar ratio from balanced equation.from balanced equation.

Step 7: Convert from moles of unknown Step 7: Convert from moles of unknown to grams of unknown using molar mass.to grams of unknown using molar mass.

Step 8: Perform calculation and express Step 8: Perform calculation and express answer in correct sig figs, unit, and answer in correct sig figs, unit, and substance.substance.

*Units must be placed so that they cancel *Units must be placed so that they cancel throughout the solving process.throughout the solving process.

25 g O25 g O22 ( 1 mol ) ( 2 mol H ( 1 mol ) ( 2 mol H22O ) ( 18.02 g O ) ( 18.02 g ))

1 ( 32.00g) ( 1 mol O1 ( 32.00g) ( 1 mol O22 ) ( 1 ) ( 1 mol ) =mol ) =

28 g H28 g H22OO

producedproduced

Giv: 25 g O2

Unk: g H2O

CC44HH1010 + O + O22 CO CO22 + H + H22O O How much (g) of carbon dioxide gas is How much (g) of carbon dioxide gas is

produced from the complete combustion produced from the complete combustion of 50.0 g butane (Cof 50.0 g butane (C44OO1010) and excess ) and excess oxygen?oxygen?

Step 1: Balance equation:Step 1: Balance equation:

2C2C44HH1010 + 13O + 13O22 8CO 8CO22 + 10H + 10H22O O Step 2: Identify given and unknown. Step 2: Identify given and unknown.

Given is 50.0 g CGiven is 50.0 g C44HH1010, Unknown is g of CO, Unknown is g of CO22

(Excess oxygen is background information (Excess oxygen is background information at this point. We will deal with it during at this point. We will deal with it during limiting and excess reactant calculations)limiting and excess reactant calculations)

Step 3:Step 3:

Given/Given/

UnknowUnknownn

AmounAmountt

MoleMoless

Molar MassMolar Mass

CC44HH1010 50.0 g50.0 g 22 58.14 58.14 g/molg/mol

COCO22 XX 88 44.01 44.01 g/molg/mol

Steps 4-8

50.0g C4H10 ( 1 mol )( 8 mol CO2 )(44.01 g)

1 (58.14g)( 2 mol C4H10)( 1 mol ) =

151 g CO2

produced

EX: How much hydrogen is produced EX: How much hydrogen is produced from the reaction of 25.5 g of from the reaction of 25.5 g of magnesium metal with excess magnesium metal with excess hydrochloric acid?hydrochloric acid?

Step 1: Balanced equation. We must Step 1: Balanced equation. We must write a write a reaction based on reactants, reaction based on reactants, predict predict products, then we can products, then we can balance.balance.

Mg + HCl Mg + HCl MgCl MgCl22 + H + H22

Balance: Balance: Mg + 2HCl Mg + 2HCl MgCl MgCl22 + H + H22

Step 2: Given is 25.5 g Mg, Unknown is g HStep 2: Given is 25.5 g Mg, Unknown is g H22

At this point, you may skip step 3 (table) At this point, you may skip step 3 (table) and solve the problems using the remaining and solve the problems using the remaining steps.steps.

Steps 4-8:Steps 4-8:

25.5 g Mg ( 1 mol ) (1 mol H25.5 g Mg ( 1 mol ) (1 mol H22 ) ( 2.02 g ) ) ( 2.02 g ) 1 (24.31g) (1 mol Mg) ( 1 mol ) =1 (24.31g) (1 mol Mg) ( 1 mol ) =

2.12 g H2.12 g H22

producedproducedGiv: 25.5 g MgUnk: g H2

Practice:Practice:

Given the following equation: Fe + SGiven the following equation: Fe + S88 FeS FeSa. What mass of iron is needed to react with 16.0 grams a. What mass of iron is needed to react with 16.0 grams of sulfur? of sulfur? b. How many grams of FeS are produced?b. How many grams of FeS are produced?

Given the following equation: NaClOGiven the following equation: NaClO33 NaCl + O NaCl + O22

a. 12.0 g of NaClOa. 12.0 g of NaClO33 will produce how many grams of O will produce how many grams of O22??b. How many grams of NaCl are produced when 80.0 b. How many grams of NaCl are produced when 80.0 grams of Ograms of O22 are produced? are produced?

Given the following equation: Cu + AgNOGiven the following equation: Cu + AgNO33 Cu(NO Cu(NO33))22 + Ag + Aga. How much copper is needed to react with 350 g of a. How much copper is needed to react with 350 g of AgNOAgNO33? ? b. If 115.5 grams of copper(II) nitrate were produced, b. If 115.5 grams of copper(II) nitrate were produced, how many grams of silver nitrate reacted?how many grams of silver nitrate reacted?

Mole Mole Mass Mass Given the equation: HGiven the equation: H22(g) + O(g) + O22(g) (g) H H22O(g)O(g)

Calculate moles of HCalculate moles of H22O produced if 25 O produced if 25 grams of oxygen is reacted with excess grams of oxygen is reacted with excess hydrogen.hydrogen.

Step 1: Start with a balanced equation.Step 1: Start with a balanced equation.

Step 2: Identify given and unknown.Step 2: Identify given and unknown.

Step 3: Place given over 1.Step 3: Place given over 1.

Step 4: Convert from grams of given to Step 4: Convert from grams of given to moles of given using molar mass.moles of given using molar mass.

Step 5: Convert from moles of given to Step 5: Convert from moles of given to moles of unknown.moles of unknown.

*You may now stop because you are in *You may now stop because you are in moles of unknown. No need to use moles of unknown. No need to use the molar mass of unknown. the molar mass of unknown.

Step 6: Perform calculation and Step 6: Perform calculation and express answer in correct sig figs, express answer in correct sig figs, unit, and substance. unit, and substance.

Step 1: 2HStep 1: 2H22(g) + O(g) + O22(g) (g) 2H 2H22O(g)O(g)

Step 2: Given is 25 g of OStep 2: Given is 25 g of O22, Unknown is , Unknown is mol of Hmol of H22OO

Step 3-5Step 3-5

25 g O25 g O22 ( 1 mol ) ( 2 mol H ( 1 mol ) ( 2 mol H22O )O )

1 ( 32.00 g ) ( 1 mol O1 ( 32.00 g ) ( 1 mol O22 ) = ) =

1.6 mol H1.6 mol H22OO

producedproduced

KClOKClO33(aq) (aq) KCl(aq) + O KCl(aq) + O22(g) (g)

Using the above reaction:Using the above reaction:

1.1. How much potassium chloride is How much potassium chloride is produced from the decomposition of produced from the decomposition of 5.0 mol of potassium chlorate?5.0 mol of potassium chlorate?

2.2. How many moles of oxygen are How many moles of oxygen are produced if 45.09 g of potassium produced if 45.09 g of potassium chloride are produced? chloride are produced?

Mol Mol Mol Mol

These are the easiest type of These are the easiest type of conversion problem. Just be careful conversion problem. Just be careful to not do extra steps!to not do extra steps!

NaNa22O + HO + H22O O NaOH NaOH How many moles of NaOH are How many moles of NaOH are

produced if 6.5 moles of sodium produced if 6.5 moles of sodium oxide is combined with excess oxide is combined with excess water?water?

Step 1: Balanced equation.Step 1: Balanced equation.

Step 2: Identify given and unknown.Step 2: Identify given and unknown.

Step 3: Place given over 1.Step 3: Place given over 1.

Step 4: Convert from moles of given to Step 4: Convert from moles of given to moles of unknown.moles of unknown.

Step 5: Perform calculation and express Step 5: Perform calculation and express answer in correct sig figs, unit, and answer in correct sig figs, unit, and substance.substance.

Step 1: NaStep 1: Na22O + HO + H22O O 2NaOH 2NaOH

Step 2: Given is 6.5 mol NaStep 2: Given is 6.5 mol Na22O, Unknown O, Unknown is mol NaOHis mol NaOH

Step 3-5:Step 3-5:

6.5 mol Na6.5 mol Na22O ( 2 mol NaOH )O ( 2 mol NaOH )

1 ( 1 mol Na1 ( 1 mol Na22O ) = 13 mol O ) = 13 mol NaOHNaOH

Given the equation:Given the equation:

CHCH44(g) + O(g) + O22(g) (g) CO CO22(g) + H(g) + H22O(g)O(g)

1.1. Calculate moles of water produced by Calculate moles of water produced by the combustion of 500.0 moles of the combustion of 500.0 moles of methane.methane.

2.2. How many moles of carbon dioxide is How many moles of carbon dioxide is produced if 5 moles of water is produced if 5 moles of water is produced?produced?

3.3. How many moles of oxygen are needed How many moles of oxygen are needed to produce 65.5 moles of carbon dioxide?to produce 65.5 moles of carbon dioxide?

Mixed Stoic ProblemsMixed Stoic Problems

1.1. How much magnesium oxide is produced How much magnesium oxide is produced from 7.5 moles of magnesium ribbon?from 7.5 moles of magnesium ribbon?

2.2. How many moles of sulfuric acid is needed to How many moles of sulfuric acid is needed to react with 100. moles of copper(II) nitrate?react with 100. moles of copper(II) nitrate?

3.3. If you combust 1500 g of propane (CIf you combust 1500 g of propane (C33HH88), ), how much carbon dioxide can be produced?how much carbon dioxide can be produced?

4.4. How many moles of aluminum is produced How many moles of aluminum is produced by the complete decomposition of 45 g of by the complete decomposition of 45 g of aluminum oxide?aluminum oxide?

5.5. How much copper is produced if zinc reacts How much copper is produced if zinc reacts with 35.5 moles of copper(I) sulfate?with 35.5 moles of copper(I) sulfate?

Limiting and Excess Limiting and Excess ReactantsReactants

In the hamburger analogy, it would take 1 bun, 2 In the hamburger analogy, it would take 1 bun, 2 hamburger patties, 2 slices cheese, and 4 strips hamburger patties, 2 slices cheese, and 4 strips bacon to make 1 bacon double cheeseburger.bacon to make 1 bacon double cheeseburger.

If given certain amount of each item, you could If given certain amount of each item, you could calculate the number of cheeseburgers you calculate the number of cheeseburgers you could produce.could produce.

You could also calculate how much of each You could also calculate how much of each substance would be left over if one of your items substance would be left over if one of your items was limited.was limited.

This is the basis for limiting reactant problems.This is the basis for limiting reactant problems. One reactant will always limit the amount of One reactant will always limit the amount of

product that can be produced in a chemical rx.product that can be produced in a chemical rx.

Refer to the diagram demonstrating Refer to the diagram demonstrating the reaction between nitrogen and the reaction between nitrogen and hydrogen.hydrogen.

1. Write the balanced equation for the 1. Write the balanced equation for the reaction.reaction.

2. Which of the reactants is limiting the 2. Which of the reactants is limiting the amount of product that is made?amount of product that is made?

3. How many molecules of this limiting 3. How many molecules of this limiting reactant is in excess?reactant is in excess?

4. How many molecules of product are 4. How many molecules of product are made?made?

How many grams of product can be How many grams of product can be produced if 50.0 grams of hydrogen produced if 50.0 grams of hydrogen combines with 50.0 grams of oxygen?combines with 50.0 grams of oxygen?

Step 1: Write a balanced equation.Step 1: Write a balanced equation.Step 2: Perform 2 mass to mass Step 2: Perform 2 mass to mass

conversions, one for each reactant.conversions, one for each reactant.Step 3: The conversion with the least Step 3: The conversion with the least

amount of product produced represents amount of product produced represents that of the limiting reactant.that of the limiting reactant.

*You now know the limiting reactant and *You now know the limiting reactant and the amount of product that can be the amount of product that can be produced. The amount of excess will be produced. The amount of excess will be calculated later.calculated later.

Step 1: 2HStep 1: 2H22(g) + O(g) + O22(g) (g) 2H 2H22OOStep 2: Step 2:

50.0 g H50.0 g H22 ( 1 mol )( 2 mol H ( 1 mol )( 2 mol H22O )(18.02 g)O )(18.02 g)

1 ( 2.02 g )( 2 mol H1 ( 2.02 g )( 2 mol H2 2 )( 1 mol ) =)( 1 mol ) =

446 g H446 g H22OOproducedproduced

50.0 g O50.0 g O22 ( 1 mol )( 2 mol H ( 1 mol )( 2 mol H22O )(18.02 g)O )(18.02 g)

1 ( 32.00 g )( 1 mol O1 ( 32.00 g )( 1 mol O22 )( 1 mol ) = )( 1 mol ) =

56.3 g H56.3 g H22OOproducedproduced

Step 3: OStep 3: O22 is the limiting reactant because is the limiting reactant because only 56.3 grams of water is produced.only 56.3 grams of water is produced.

Calculate grams of copper produced if 25 Calculate grams of copper produced if 25 grams of zinc is reacted with 50. grams grams of zinc is reacted with 50. grams of copper(I) carbonate.of copper(I) carbonate.

Step 1: Zn + CuStep 1: Zn + Cu22COCO33 ZnCO ZnCO33 + 2Cu + 2Cu

Step 2:Step 2:

25 g Zn ( 1 mol )( 2 mol Cu )( 63.55 g )25 g Zn ( 1 mol )( 2 mol Cu )( 63.55 g )

1 ( 65.37 g)( 1 mol Zn )( 1 mol ) =1 ( 65.37 g)( 1 mol Zn )( 1 mol ) =

49 g Cu produced49 g Cu produced

50.g 50.g CuCu22COCO3 3 ( 1mol )( 2 mol Cu )( 63.55 g )( 1mol )( 2 mol Cu )( 63.55 g )

1 (187.11g)(1mol 1 (187.11g)(1mol CuCu22COCO33)( )( 1 mol ) 1 mol ) ==

34 g Cu produced34 g Cu produced

50. grams of cuprous carbonate is the 50. grams of cuprous carbonate is the limiting reactant.limiting reactant.

You can also calculate moles of You can also calculate moles of product produced from given product produced from given quantities of reactants.quantities of reactants.

-Just convert grams of each reactant -Just convert grams of each reactant into moles. The one with the least into moles. The one with the least amount of moles is the limiting amount of moles is the limiting reactant.reactant.

Excess ReactantExcess Reactant In order to calculate how much excess In order to calculate how much excess

reactant remains, follow these steps:reactant remains, follow these steps:Step 1: Find the limiting reactant.Step 1: Find the limiting reactant.Step 2: Convert from grams of limiting Step 2: Convert from grams of limiting

reactant to grams of the other reactant. reactant to grams of the other reactant. This is the amount of that reactant This is the amount of that reactant consumedconsumed..

Step 3: Subtract the amount of other reactant Step 3: Subtract the amount of other reactant consumed from the original quantity in the consumed from the original quantity in the problem to find the amount in excess.problem to find the amount in excess.

* Original grams – consumed grams = grams * Original grams – consumed grams = grams excess excess

How much of the excess reactant remains How much of the excess reactant remains if 100. g of bromine reacts with 75.0 if 100. g of bromine reacts with 75.0 grams of magnesium ribbon?grams of magnesium ribbon?

Step 1: BrStep 1: Br22(g) + Mg(s) (g) + Mg(s) MgBr MgBr22

100. g Br100. g Br22 ( 1 mol ) ( 1 mol MgBr ( 1 mol ) ( 1 mol MgBr22) )

1 ( 159.80 g) ( 1 mol Br1 ( 159.80 g) ( 1 mol Br22 ) = ) =

0.626 mol MgBr0.626 mol MgBr22

75.0 g Mg ( 1 mol ) ( 1 mol MgBr75.0 g Mg ( 1 mol ) ( 1 mol MgBr22 ) )

1 ( 24.31 g ) ( 1 mol Mg ) =1 ( 24.31 g ) ( 1 mol Mg ) =

3.09 mol MgBr3.09 mol MgBr22

*Br*Br22 is the limiting reactant. is the limiting reactant.

Step 2: Step 2:

100. g Br100. g Br22 ( 1 mol )( 1 mol Mg)( 24.31 g) ( 1 mol )( 1 mol Mg)( 24.31 g)

1 ( 159.80g)( 1 mol Br1 ( 159.80g)( 1 mol Br22)( 1 mol ) )( 1 mol ) ==

15.2 g Mg15.2 g Mg consumedconsumed

Step 3:Step 3:

75.0 g - 15.2 g = 59.8 g Mg 75.0 g - 15.2 g = 59.8 g Mg excessexcess

Original

Quantity

From

Problem

Quantity

Consumed

From

Calculation

Calculate grams of product and grams Calculate grams of product and grams of excess reactant if 80. g of aluminum of excess reactant if 80. g of aluminum reacts with 80. g of chlorine.reacts with 80. g of chlorine.

Your Challenge:Your Challenge:

If given 25.0 grams of each reactant in If given 25.0 grams of each reactant in the equation:the equation:

2CH2CH22CHCHCHCH33 + 2NH + 2NH33 + 3O + 3O22 2CH 2CH22CHCN + 6HCHCN + 6H22OO 1.1. Find the limiting reactant.Find the limiting reactant.2.2. Calculate grams of each product Calculate grams of each product

produced.produced.3.3. Calculate grams of each excess reactant.Calculate grams of each excess reactant.*Hint: Follow steps for finding the limiting *Hint: Follow steps for finding the limiting

reactant. Just do 3 mass to mass reactant. Just do 3 mass to mass problems instead of 2.problems instead of 2.

Theoretical Yield and Percent Theoretical Yield and Percent YieldYield

Thus far in all our calculations we assumed that Thus far in all our calculations we assumed that the reaction conditions were ideal and led to the reaction conditions were ideal and led to reactions that went to 100% completion. reactions that went to 100% completion.

Calculation of product mass with these ideal Calculation of product mass with these ideal conditions in mind are known as the conditions in mind are known as the ""theoretical yieldtheoretical yield". ".

However, in reality reaction are influenced by However, in reality reaction are influenced by many "external" factors (i.e., factors apart from many "external" factors (i.e., factors apart from the reactants). Some of these factors are:the reactants). Some of these factors are:

*Temp, Pressure, Accuracy of Equipment*Temp, Pressure, Accuracy of Equipment

You have been calculating the theoretical You have been calculating the theoretical yield for every stoic problem so far.yield for every stoic problem so far.Calculate theoretical yield of product if 25 Calculate theoretical yield of product if 25 g of hydrogen combines with 25 g of g of hydrogen combines with 25 g of oxygen.oxygen.

Step 1: Find the limiting.Step 1: Find the limiting.Step 2: Identify that the least amount of Step 2: Identify that the least amount of

product produced not only indicates your product produced not only indicates your limiting reactant, it indicates the limiting reactant, it indicates the theoretical yield of product that could be theoretical yield of product that could be produced if the experimental conditions produced if the experimental conditions were 100% accurate.were 100% accurate.

2H2H22(g) + O(g) + O22(g) (g) 2H 2H22OO

25 g H25 g H22 ( 1 mol ) ( 2 mol H ( 1 mol ) ( 2 mol H22O ) (18.02 g )O ) (18.02 g )

1 (2.02 g ) ( 2 mol H1 (2.02 g ) ( 2 mol H22 ) ( 1 mol ) = ) ( 1 mol ) =

220 g H220 g H22O producedO produced

25 g O25 g O22 ( 1 mol ) ( 2 mol H ( 1 mol ) ( 2 mol H22O ) ( 18.02 g )O ) ( 18.02 g )

1 (32.00g) ( 1 mol O1 (32.00g) ( 1 mol O22 ) ( 1 mol ) = ) ( 1 mol ) =

28 g H28 g H22O producedO produced

*Oxygen is the limiting reactant and 28 g of *Oxygen is the limiting reactant and 28 g of HH22O is the theoretical yield of product that O is the theoretical yield of product that can be made if everything went 100%.can be made if everything went 100%.

Percent YieldPercent Yield In most cases the reaction In most cases the reaction does notdoes not go go

to completion. to completion. In this case the mass of products In this case the mass of products

formed (the actual yield) is less than formed (the actual yield) is less than the theoretical yield. the theoretical yield.

A quantity that describes this less-A quantity that describes this less-than-ideal yield is known as the than-ideal yield is known as the ""percent yieldpercent yield": ":

The actual amount is either given in the The actual amount is either given in the problem or obtained through problem or obtained through experimentation.experimentation.

In the previous reaction between In the previous reaction between hydrogen and oxygen, an hydrogen and oxygen, an experiment experiment yielded 25.6 gramsyielded 25.6 grams of water. of water. Calculate the percent yield of the Calculate the percent yield of the product.product.

% yield = % yield = 25.6 g25.6 g x 100 = x 100 =

28 g28 g

91% yield

50. g of silver nitrate is mixed with 50. 50. g of silver nitrate is mixed with 50. g of hydrochloric acid in a water based g of hydrochloric acid in a water based solution. A white precipitate forms solution. A white precipitate forms (silver chloride). The solution is filtered (silver chloride). The solution is filtered and the white precipitate collected and and the white precipitate collected and dried. The dried precipitate is dried. The dried precipitate is measured to have a mass of measured to have a mass of 40.3 g40.3 g. . What is the theoretical and percent What is the theoretical and percent yield?yield?