iii. stoichiometry stoy – kee – ahm –eh - tree chapter 12

III. StoichiometryStoy – kee – ahm –eh - tree

Chapter 12

SectionsClick the section to jump to the slides

•Mole Ratios•Mole-to-Mole Calculations•Mole-to-Mass Calculations•Particle Calculations•Molar Volume Calculations•Limiting Reactants •Percent Yields

Things you should remember

• From the Moles Unit:– Identify particles as atoms, molecules (mc),

and formula units (fun)– 1 mole = 6.02 x 1023 atoms, molecules, or

formula units– 1 mole substance = mass (in grams) from the

periodic table• From the Naming & Formulas Unit:

– How to write a formula given a chemical name• From the Chemical Reactions Unit:

– How to write a chemical equation given words– Balancing equations

I. Stoichiometry

o stoikheion, meaning element

o metron, meaning measureo Thus Stoichiometry-

measuring elements!

III. Stoichiometry

oStoichiometry (sometimes called reaction stoichiometry to specify its use in analyzing a chemical reaction) is the calculation of quantitative (numbers) relationships between the reactants and products in a balanced chemical reaction.

A. Basis for Calculations

•The basis for properly working stoichiometry problems is the Balanced Chemical Equation and the Mole Ratio. These are VITAL TO YOUR SURVIVAL IN STOICHIOMETRY!!!

1. Review of a Balanced Chemical Equation

• You must always check to ensure you have a proper chemical equation. It is not correctly balanced you can not use it for any calculations!!!!

Ex: Properly Balance the following equation

__H2SO4(aq) + __NaHCO3(s)

__Na2SO4(aq) + __H2O(l) + __CO2(g)

2

2 2

2. Molar Ratio

• Once the equation is properly balanced, there are relationships between all of the compounds involved. These are called MOLE RATIOS.

• any two compounds can be written as a relationship in terms of moles

What’s the relationship between NaHCO3 and Na2SO4?

2 mole NaHCO3 = 1 mol Na2SO4

(it’s just the coefficients!!)

What’s the relationship between H2SO4 and CO2?

1 mol H2SO4 = 2 mol CO2

(it’s just the coefficients!!)

For the Reaction:H2SO4 + 2NaHCO3 Na2SO4 + 2H2O +

2CO2

• 1 mol H2SO4 = 2 mol NaHCO3

or

• 1 mc H2SO4 = 2 mc NaHCO3

or

• 1 mol/mc H2SO4

2 mol/mc NaHCO3

or

• 2 mol/mc NaHCO3

1 mole/mc H2SO4


2CO2

• 1 mol H2SO4 = 2 mol CO2

or

• 1 mc H2SO4 = 2 mc CO2

or

• 1 mol/mc H2SO4

2 mol/mc CO2

or

• 2 mol/mc CO2

1 mole/mc H2SO4


2CO2

Example: Iron reacts with oxygen to create iron(III) oxide.

After writing a balanced equation, write down 3 possible relationships. (hint: just look at the coefficients.)

Skeletal equation:

Balanced equation:

Fe + O2 Fe2O3

4Fe + 3O2 2Fe2O3

Relationships: 4 mol Fe = 3 mol O2

4 mol Fe = 2 mol Fe2O3

3 mol O2 = 2 mol Fe2O3

Practice1. Write three mole ratios (relationships)

from the reaction below:

Al2S3 + H2O Al(OH)3 + H2S1 mol Al2S3 = 6 mol H2O

26 3

1 mol Al2S3 = 2 mol Al(OH)3

1 mol Al2S3 = 3 mol H2S6 mol H2O = 2 mol Al(OH)3

6 mol H2O = 3 mol H2S2 mol Al(OH)3 = 3 mol H2SY

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Practice1. Write three mole ratios (relationships)

from the reaction below:Al2S3 + H2O Al(OH)3 + H2S

1 mol Al2S3 = 6 mol H2O

26 3

1 mol Al2S3 = 2 mol Al(OH)3

1 mol Al2S3 = 3 mol H2S6 mol H2O = 2 mol Al(OH)3

6 mol H2O = 3 mol H2S

Practice2. Aluminum is produced by decomposing

aluminum oxide into aluminum and oxygen.a. Write a balanced equation.

b. Write all the molar ratios that can be derived from this equation.

2Al2O3 4Al + 3O2

2 mol Al2O3 = 4 mol Al 2 mol Al2O3 = 3 mol O2

4 mol Al = 3 mol O2

NOTE:

• NOTE: Every time you do a stoichiometric calculation you MUST use a mole ratio. The mole ratio allows you to compare one compound in an equation with another. Don’t forget to balance your chemical equations!!!

B. Calculating Problems

B. Calculating Problems - Stoich it up!

• Before any stoich problem you have to set it up. Consider this the pre-game warm-up. This should become second nature to you.

Pre-game Warm-up:1. Write a balanced reaction.2. Determine your given & want3. Determine your relationships. If you see…• 2 different substances, determine their mole

ratio• mass (g, mg, kg), calculate the molar mass

of that substance• atoms, molecules, or fun, 1 mol = 6.02 x 1023

of that type• extras like mg or kg, you know what to do

Now you are ready to solve- IT’S GAME TIME.

Game Time:

1. put your GIVEN OVER 12. place your relationships where the

units cancel out diagonally3. everything equal to each other goes

above and below each other4. cancel out your units until you are

left over with your wanted

Here is a flow chart that we will dissect this unit to do

our problems

For most of the examples we will be using this equation:

N2 (g) + 3H2 (g) 2NH3 (g)

Haber Process: an industrial process for producing ammonia from nitrogen and

hydrogen by combining them under high pressure in the present of an iron catalyst

source: worldnet.princeton.edu

Haber Process:

1. Moles to Moles

Ex 1: How many moles of ammonia are produced from 4.00 moles of hydrogen

gas in the presence of excess nitrogen?

Balanced Equation:N2 (g) + 3H2 (g) 2NH3 (g)

1

4.00 mol H2

3 mol H2

2.67

Given :Want :Relationships:

4.00 mol H2

? mol NH3

2 mol NH3 = 3 mol H2

= mol NH32 mol NH3x

No grams, No molar mass!!

No fun, mc, No 6.02 x 1023!!

= mol N2

Ex 2: How many moles of nitrogen were used if 7.8 moles of ammonia were made

in excess hydrogen gas?


1 mol N2

1

7.8 mol NH3

2 mol NH3

3.9


7.8 mol NH3

? mol N2

2 mol NH3 = 1 mol N2

X


No fun, mc, No 6.02 x 1023!!

= moles H2

Ex 3: How many moles of hydrogen react with 13 moles of nitrogen?


1

13 mol N2

1 mol N2

39


13 mol N2

? mol H2

1 mol N2 = 3 mol H2

3 mol H2x


No fun, mc, No 6.02 x 1023!!

2. Moles to Mass/ Mass to Moles

Steps to Success!1. Write a balanced reaction.2. Determine your given & want3. Determine your relationships. If you

see…– 2 different substances, determine their

mole ratio– mass (g, mg, kg), calculate the molar

mass of that substance– atoms, molecules, or f.un, 1 mol = 6.02 x

1023 of that type– extras like mg or kg, you know what to do

Ex 1: How many grams of ammonia are produced from 4.00 moles of hydrogen

gas in excess nitrogen?



4.00 mol H2

? g NH3


How do I know when to use mole ratios,

molar mass or 6.02 x 1023?

Look at the “given” and “want” for

clues

1 mol NH3 = 17.031g NH3No fun, mc, No 6.02 x 1023!!

= g NH3

Ex 1: How many grams of ammonia are produced from 4 moles of hydrogen gas in

excess nitrogen?

1

4.00 mol H2

3 mol H2

45.4

2 mol NH3x



4.00 mol H2

? g NH3

2 mol NH3 = 3 mol H21 mol NH3 = 17.031g

NH3

1 mol NH3

17.031g NH3x

Ex 2: How many moles of ammonia are produced from 4.00 grams of hydrogen gas in

excess nitrogen?



4.00 g H2

? mol NH3


Remember your Steps

to Success!

1 mol H2 = 2.016g H2

No fun, mc, No 6.02 x 1023!!

= mol NH3

Ex 2: How many moles of ammonia are produced from 4 grams of hydrogen gas in

excess nitrogen?

1

4.00 g H2

2.016g H2

1.32

1 mol H2

x


3 mol H2

2 mol NH3x


4.00 g H2

? mol NH3


1 mol H2 = 2.016g H2

Practice1. How many grams of nitrogen will

react with 3.40 moles of hydrogen to produce ammonia?

2. How many grams of hydrogen are required to make 54.0 moles of ammonia in excess nitrogen?

3. How many moles of nitrogen react completely with 3.70 moles of hydrogen?

31.7 g N2

163 g H2

1.23 mol N2

3. Mass to Mass

Ex 1: How many grams of ammonia are produced from 4.00 grams of hydrogen

gas in excess Nitrogen?



4.00 g H2

? g NH3


Remember your Steps

to Success!

1 mol H2 = 2.016g H2

1 mol NH3 = 17.031 g NH3

= g NH3

Ex 1: How many grams of ammonia are produced from 4 grams of hydrogen gas in

excess Nitrogen?

1

4.00 g H2

2.016g H2

22.5

1 mol H2x


3 mol H2

2 mol NH3x


4.00 g H2

? g NH3


1 mol H2 = 2.016g H2

1 mol NH3 = 17.031g NH3

1 mol NH3

17.031g NH3x

Ex 2: How many grams of Nitrogen react with 12.0 grams of Hydrogen?



12.0 g H2

? g N2

Remember your Steps

to Success!

1 mol H2 = 2.016g H2

1 mol N2 = 28.014 g N2

= g N2

Ex 2: How many grams of Nitrogen react with 12.0 grams of Hydrogen?

112.0 g H2

2.016g H2

55.6

1 mol H2x


3 mol H2

1 mol N2x


12.0 g H2

? g N2

1 mol N2 = 3 mol H2

1 mol H2 = 2.016g H2

1 mol N2 = 28.014 g N2

1 mol N2

28.014g N2

x

Practice

1. Determine the mass of NH3 produced from 280 g of N2.

2. What mass of nitrogen is needed to produce 100. kg of ammonia?

340 g NH3

8.22 x 104 g N2

3. MC/F.UN/ Atoms to Mass

Ex 1: How many grams of ammonia are produced from 2.09 x 1014 mc

of hydrogen gas?Balanced Equation:

N2 (g) + 3H2 (g) 2NH3 (g)


2.09 x 1014 mc H2

? g NH3


1 mol NH3 = 17.031g NH31 mol H2 = 6.02 x 1023 mc

H2

Ex 1: How many grams of ammonia are produced from 2.09 x 1014 mc

of hydrogen gas?Balanced Equation:

N2 (g) + 3H2 (g) 2NH3 (g)


2.09 x 1014 mc H2

? g NH3


1 mol NH3 = 17.031g NH3

1 mol H2 = 6.02 x 1023 mc H2

= g NH3

1

2.09 x 1014 mc H2

6.02 x 1023 mc H2

3.94 x 10-9

1 mol H2x3 mol H2

2 mol NH3x1 mol NH3

17.031g NH3x

Ex 2: How many molecules of ammonia are produced from 40.2 g of H2 in the

presence of excess N2?Balanced Equation:

N2 (g) + 3H2 (g) 2NH3 (g)


40.2g H2

? mc NH3


1 mol H2 = 2.016g H2

1 mol NH3 = 6.02 x 1023 mc NH3

Ex 2: How many molecules of ammonia are produced from 40.2 g of

H2 in the presence of excess N2?

= mc NH3

1

40.2g H2

2.016g H2

8.00 x 1024

1 mol H2x3 mol H2

2 mol NH3x1 mol NH3

6.02 x 1023 mc NH3

x



40.2g H2

? mc NH32 mol NH3 = 3 mol H2

1 mol H2 = 2.016g H2

1 mol NH3 = 6.02 x 1023 mc NH3

Practice

1. How many mc of nitrogen will react with exactly 7.04 grams of hydrogen?

2. How many mc of ammonia will 2.09 x 1021 mc of N2 produce in excess hydrogen?

7.01 x 1023 mc N2

4.18 x 1021 mc NH3

1) How many mc of nitrogen will react with exactly 7.04 grams of

hydrogen?

= mc N2

1

7.04 g H2

2.016g H2

7.01 x 1023

1 mol H2x3 mol H2

1 mol N2x1 mol N2

6.02 x 1023 mc N2x



7.04 g H2

? mc N21 mol N2 = 3 mol H2

1 mol H2 = 2.016g H2

1 mol N2 = 6.02 x 1023 mc N2

How many mc of ammonia will 2.09 x 1021 mc of N2 produce in

excess hydrogen?Balanced Equation:

N2 (g) + 3H2 (g) 2NH3 (g)


2.09 x 1013 mc N2

? mc NH3

2 mol NH3 = 1 mol N2

1 mol NH3 = 6.02 x 1023 NH3

1 mol N2 = 6.02 x 1023 mc N2

= mc NH3

1

2.09 x 1013 mc N2

6.02 x 1023 mc N2

4.18 x 1021

1 mol N2x1 mol N2

2 mol NH3x

1 mol NH3

6.02 x 1023 mc NH3x

4. Molar Volume of Gases (Liters)

4. Molar Volume of Gases (liters)

• There is one new conversion: 1 mol of an ideal gas = 22.4 L of an ideal gas. For now we will assume every gas is an ideal gas at STP. You follow the same steps we have been following all along to solve this problem. In a reaction involving gases, Avogadro found that their volumes combine in whole number ratios. In other words, the coefficients in the chemical reaction also represent volume.

For gases:When you see liters…

1 mol __ = 22.4 L __

Ex 1: How many moles of nitrogen gas are needed to react with 44.8 liters of hydrogen

gas to produce ammonia gas?



44.8 L H2

? mol N2

1 mol N2 = 3 mol H2

Remember your Steps

to Success!

1 mol H2 = 22.4 L H2

No grams, No molar mass

No fun, mc, No 6.02 x 1023!!

= mol N2

Ex 1: How many moles of nitrogen gas are needed to react with 44.8 liters of hydrogen

gas to produce ammonia gas?

1

44.8 L H2

22.4 L H2

0.667 1 mol H2

x


3 mol H2

1 mol N2x


44.8 L H2

? mol N2

1 mol N2 = 3 mol H2

1 mol H2 = 22.4 L H2

Given over 1 1 mol = 22.4 L Mole ratio

Ex 2: If 5.00 moles of H2 react with excess N2, how many liters of NH3 are

produced? N2 (g) + 3H2 (g) 2NH3 (g)


5.00 mol H2

? L NH3


Remember your Steps

to Success!

1 mol NH3 = 22.4 L NH3

No grams, No molar mass

No fun, mc, No 6.02 x 1023!!

= L NH31

5.00 mol H2

3 mol H2

74.7 2 mol NH3x 1 mol NH3

22.4 L NH3x

Given over 1 Mole ratio 1 mol = 22.4 L

Ex 2: If 5.00 moles of H2 react with excess N2, how many liters of NH3 are

produced? N2 (g) + 3H2 (g) 2NH3 (g)


5.00 mol H2

? L NH3


1 mol NH3 = 22.4 L NH3

II. Limiting Reactants and

Excess Reactants

• In reality, a scientist does not always add chemicals in perfect proportions. There is usually one reactant that is in “excess”.

• The limiting reactant or limiting reagent is the reactant that limits the amounts of the other reactants that can combine and the amount of product that can form in a chemical reaction. “which reactant are

you going to run out of first?”

• The substance that is not used up completely in a reaction is sometimes called the excess reactant (excess)

• Once one of the reactants is used up, no more product can be formed

“which reactant do you have extra of?”

+ 4

4 13Which is the limiting reactant

4 shells

13 tires

1 shell

4 tires

1 car

1 car = 4 cars can be made if 4 shells are available

= 3.25 cars can be made if 13 tires are available

Less product can be produced, therefore tires must be the limiting reactant

Activity

• Write a recipe for the perfect burger

Ex. Consider the following balanced equation: Zn(s) + 2HCl(aq)

ZnCl2(aq) + H2(g)If 1.21 moles of zinc are added to 2.64

moles of HCl, which reactant is in excess, and which is the limiting

reactant? 1.21 mol Zn

2.64 mol HCl

Available

2 mol HCl

2 mol HCl1 mol Zn

1 mol Zn

= 2.42 mol HCl

= 1.32 mol Zn

Needed

Ex. Consider the following balanced equation: Zn(s) + 2HCl(aq)

ZnCl2(aq) + H2(g)If 1.21 moles of zinc are added to 2.64

moles of HCl, which reactant is in excess, and which is the limiting

reactant? Available VS Needed

1.21 mol Zn 1.32 mol Zn

2.64 mol HCl 2.42 mol HCl

<

>

Limiting reactant

Excess reactant

To find the limiting reactant and the excess

reactant:1) Determine the balanced chemical equation2) Convert the given/available amounts reactants to

the number of moles of the other reactant(s)3) Compare the available amounts to the needed

amounts:*If the available amount > needed amount, it is excess

reactant

* If the available amount < needed amount, it is the limiting reactant

4) To calculate the amount of excess:-Available Amount – Needed Amount = Amount of Excess

5) To calculate the amount of product produced:-G: amount of available limiting reactant-W: amount of product

ExampleSome rocket engines use a mixture

of hydrazine, N2H4, and hydrogen peroxide, as the propellant. The

products formed are nitrogen gas and steam. Which is the limiting

reactant when 0.75 mol of N2H4 is mixed with 0.50 mol of H2O2? How much excess reactant, in moles,

remains unchanged? How much of each product, in moles, is formed?

Some rocket engines use a mixture of hydrazine, N2H4, and hydrogen peroxide, as the propellant. The

products formed are nitrogen gas and steam.

Step 1: Write a balanced reaction

N2H4 + H2O2 N2(g) + H2O(g)42

Which is the limiting reactant when 0.750 mol of N2H4 is mixed with 0.500 mol of

H2O2?

N2H4 + H2O2 N2(g) + H2O(g)42

0.75 mol N2H4

0.50 mol H2O2

Available

2 mol H2O2

2 mol H2O2

1 mol N2H4

1 mol N2H4

= 1.50 mol H2O2

= 0.25 mol N2H4

Needed

Excess

Limiting

Reactant


reactant:4. To calculate the amount of excess:

-Available Amount – Needed Amount = Amount of Excess

How much excess reactant, in moles, remains unchanged?

0.75 mol N2H4

0.50 mol H2O2

Available

2 mol H2O2

2 mol H2O2 =

1 mol N2H4

1 mol N2H4 =

1.50 mol H2O2

0.25 mol N2H4

Needed

Excess

*Available – Needed = Remaining excess

- = 0.50 mol N2H4


reactant:5. To calculate the amount of product

produced:-G: amount of available limiting reactant-W: amount of product

How many moles of product are formed?

0.50 mol H2O2

0.50 mol H2O2

G:W:R:

2 mol H2O2

2 mol H2O2 = 1 mol N2

mole N2

1 mol N2 = 0.25 mol N2

N2H4 + H2O2 N2(g) + H2O(g)42G: amount of available limiting reactantN2 is the first product

0.50 mol H2O2

0.50 mol H2O2

G:W:R:

2 mol H2O2

2 mol H2O2 = 4 mol H2Omole H2O

4 mol H2O = 1.0 mol H2O

G: amount of available limiting reactantH2O is the second product

Ex. Zinc metal and solid sulfur (S8) react to form solid zinc sulfide. If 2.0 moles of zinc are heated

with 2.0 mol S8, identify the limiting reactant. How many moles of excess reactant remain.

How many moles of product are formed?

2.0 mol Zn

1.0 mol S8

Available

1 mol S8

1 mol S8

8 mol Zn

8 mol Zn

= 0.25 mol S8

= 8.0 mol Zn

Needed

Balanced Equation: 8 Zn + S8 8 ZnS

Ex. Zinc metal and solid sulfur (S8) react to form solid zinc sulfide. If 2.0 moles of zinc are

heated with 1.0 mol S8, identify the limiting reactant. How many moles of excess reactant

remain. How many moles of product are formed?

Available VS Needed

2.0 mol Zn 8.0 mol Zn

1.0 mol S8 0.25 mol S8

<

>

Limiting reactant

Excess reactant

Ex. Zinc metal and solid sulfur (S8) react to form solid zinc sulfide. If 2.0 moles of zinc are



How much excess is present?(Available Amount – Needed Amount = Amount of Excess)

1.0 mol S8 – 0.25 mol S8 = 0.75 mol S8 in excess




5. To calculate the amount of product produced:-G: amount of available limiting

reactant-W: amount of product

Ex. Zinc metal and solid sulfur (S8) react to form solid zinc sulfide. If two moles of zinc are



2 mol Zn

2 mol Zn

G:W:R:

8 mol Zn

8 mol Zn = 8 mol ZnSmol ZnS

8 mol ZnS = 2 mol ZnS formed

How much product was formed?Balanced Equation: 8 Zn + S8 8 ZnS

Practice1. Carbon reacts with steam at high temperatures

to produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, identify the limiting reactant. How many moles of each product are formed?

2.4 mol C

3.1 mol H2O

Available

1 mol H2O

1 mol H2O1 mol C

1 mol C

= 2.4 mol H2O

= 3.1 mol C

Needed

Balanced Equation: C + H2O H2 + CO

Practice 1. Carbon reacts with steam at high temperatures to produce hydrogen and

carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, identify the

limiting reactant. How many moles of each product are formed?

Available VS Needed

2.4 mol C 3.1 mol C

3.1 mol H2O 2.4 mol H2O

<

>

Limiting reactant

Excess reactant




5. To calculate the amount of product produced:-G: amount of available limiting

reactant-W: amount of product

Carbon reacts with steam at high temperatures to produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam,

identify the limiting reactant. How many moles of each product are formed?

2.4 mol CG:W:R: 1 mol C = 1 mol H2

mole H2

G: amount of available limiting reactantH2 is the first product

2.4 mol CG:W:R: 1 mol C = 1 mol CO

mole COG: amount of available limiting reactantCO is the second product

Balanced Equation: C + H2O H2 + CO

2.4 mol C 1 mol H2

1 mol C= 2.4 mol H2

2.4 mol C

1 mol C

1 mol CO = 2.4 mol CO

Set VI: Reactions

2) Zn + Pb(NO3)2 Pb + Zn(NO3)2

3) Fe + 2HCl H2 + FeCl2

III. Theoretical yield, Actual yield and Percent

yield

C. Theoretical yield, Actual yield and Percent yield

• The yield of a chemical reaction is the quantity of product one obtains from a given ratio of reactants.

• The actual yield of a chemical reaction is the mass of the compound that you actually recover when you are done with the reaction.

• The actual yield is also referred to as the experimental yield

• The theoretical yield is the mass of compound you should obtain (theoretically) if everything goes perfectly. In all of the examples above we have been pretending that everything is perfect. All of our wanteds have been theoretical.

Ex 1: What is the theoretical yield of Na2SO4 in grams if 35 moles of NaOH is

reacted with sufficient H2SO4?

• Yield means product• Theoretical yield means

mathematically, what should you get? (this is why we do stoichiometry calculations)

Ex 1: What is the theoretical yield of Na2SO4 in grams if 3.50 moles of NaOH is

reacted with sufficient H2SO4?

= g Na2SO4

1

3.50 mol NaOH

2 mol NaOH

249

1 mol Na2SO4

x1 mol Na2SO4

142.042g Na2SO4x

Given over 1 Mole ratio 1 mol = 142.042g

2NaOH + H2SO4 Na2SO4 + 2H2OG:

W:

R:

3.50 mol NaOH? g Na2SO4

2 mol NaOH = 1 mol Na2SO41 mol Na2SO4 = 142.042g Na2SO4

2 different substancesYou see g Na2SO4

If they give you the actual yield and you figure out the

theoretical yield, you can find the percent yield.

Actual YieldTheoretical Yield

X 100 = % Yield

Actual YieldTheoretical Yield

= % Yield 100

Same as…

Your percent yield should not be greater than 100 because the

theoretical yield is the MAXIMUM yield you can have.

Example: From the examples above, if 221 grams of sodium sulfate were

actually collected, what is the percent yield?

221 g NaSO4 249 g NaSO4

X 100 = 88.8%

From the experiment

From the calculation

Practice1.A student conducts a single displacement

reaction that produces 2.75 grams of copper. Mathematically he determines that 3.150 grams of copper should have been produced. Calculate the student's percentage yield.

2.A student completely reacts 5.00 g of magnesium with an excess of oxygen to produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide. What is the student's percentage yield? (*Hint: you must first find the theoretical yield.)

87.3%

97.7%

Practice3. Consider the reaction below of zinc and

hydrochloric acid to produce zinc chloride and hydrogen gas.

Zn + 2HCl H2 +ZnCl2

If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl2? What is the actual yield in grams if the percent yield is 87%.

Actual yield: 93.7 g

Practice 11.A student conducts a single displacement

reaction that produces 2.75 grams of copper. Mathematically he determines that 3.15 grams of copper should have been produced. Calculate the student's percentage yield.

2.75 g Cu3.15 g Cu

Actual from the experiment

calculated

X 100 = 87.3 %

Practice 2A student completely reacts 5.00g of magnesium

with an excess of oxygen to produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide. What is the student's percentage yield? (*Hint: you must first find the theoretical yield.)

Actual yield

Theo.X 100 = % yieldactual

To calculate the theoretical yield

Theoretical yield= g MgO

1

5.00 g Mg 24.305 g Mg

8.29

1 mol Mgx2 mol Mg2 mol MgOx

2Mg + O2 2 MgOG:

W:

R:

5.00 g Mgg MgO

2 mol Mg = 2 mol MgO

1 mol MgO = 40.304 g MgO

2 different substancesYou see g Mg and g MgO

1 mol Mg = 24.305 g Mg

1 mol MgO

40.304 g MgOx

Practice 2A student completely reacts 5.00g of magnesium

with an excess of oxygen to produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide. What is the student's percentage yield? (*Hint: you must first find the theoretical yield.)

Actual yield

8.29 g MgOX 100 = 97.7%8.10 g MgO

Practice 33. Consider the reaction below of zinc and

hydrochloric acid to produce zinc chloride and hydrogen gas.

Zn + 2HCl H2 +ZnCl2

If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl2? What is the actual yield in grams if the percent yield is 87%.

Actual yield: 93.7 g

If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of

ZnCl2? What is the actual yield in grams if the percent yield is 87%.

Theoretical yield= g ZnCl2

1

58.0 g HCl

36.461 g HCl

108

1 mol HClx

2 mol HCl

1 mol ZnCl2x

Zn + 2HCl H2 +ZnCl2G:

W:

R:

58.0 g HClg ZnCl2

2 mol HCl = 1 mol ZnCl2

1 mol ZnCl2 = 136.286 g ZnCl2

2 different substancesYou see g HCl & g ZnCl2

1 mol HCl = 36.461 g HCl

1 mol ZnCl2

136.286 g ZnCl2

x

actual

Practice 3• If a sufficient mass of zinc metal combines

with 58.0 grams of HCl, what is the theoretical yield of ZnCl2? What is the actual yield in grams if the percent yield is 87%.

93.7 g108 g

Actual from the experiment

calculated

X 100 = 87 %

iii. stoichiometry stoy – kee – ahm –eh - tree chapter 12

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