stoichiometry tutorial

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STOICHIOMETRY STOICHIOMETRY TUTORIALTUTORIALRevised PPT presentation by Paul GillettiRevised PPT presentation by Paul Gilletti

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Instructions: This is a work along tutorial. Each time you click the mouse or touch the space bar on your computer, one step of the problem solving occurs. Pressing the PAGE UP key will backup the steps.

Get a pencil and paper, a periodic

table and a calculator, and

let’s get to work.

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(1-2-3) General Approach For Problem Solving:

1. Clearly identify the Unknown (the goal, or what you are trying to find) and the UNITS involved.

2. Determine what is Given and the UNITS.

3. Use conversion factors (which are really ratios) and their UNITS to CONVERT what is given into what is desired.

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Table of Contents: Click on each tab to view problem types.

Sample problem 1

Sample problem 2

Converting grams to moles

Mole to Mole Conversions

View Complete Slide Show

Gram-Mole and Gram-Gram Problems

Limiting/Excess/ Reactant and Theoretical Yield Problems :

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Sample problem for general problem solving.

Sam has entered into a 10 mile marathon. Use ALL of the following conversions (ratios) to determine how many inches there are in the race.5280 ft = 1 mile; 12 inches = 1 ft

1. What is the unknown and what units are needed?

Goal = ______ inches

2. What is given and its units?

10 miles

3. Convert using factors (ratios).

10 miles = inches

mile 1

ft 5280

ft 1

inches 12 633600

Units match

Given GoalConvert

Menu

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Sample problem #2 on problem solving.

A car is traveling at a speed of 45 miles per hr (45 miles/hr). Determine its speed in kilometers per second using the following conversion factors (ratios). 1 mile = 5280 ft; 1 ft = 12 in; 1 inch = 2.54 cm; 1 km = 1 x 103 m; 1 cm = 1 x 10-2 m; 1 hr =60 min; 1 min = 60 s

= km shr

mi 45mi

ft 5280

ft 1

in 12

in 1

cm 2.54cm 100

m 1

m10

km3 min 60

hr

s 60

min0.020

Units Match!

UnknownGiven

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Converting grams to moles.

Determine how many moles there are in 5.17 grams of Fe(C5H5)2.

Unknown

= moles Fe(C5H5)2

Given

5.17 g Fe(C5H5)2

Use the molar mass to convert grams to

moles.

Fe(C5H5)2

2 x 5 x 1.001 = 10.012 x 5 x 12.011 = 120.11

1 x 55.85 = 55.85

mol

g 185.97

g 185.97

mol0.0278

units match

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Stoichiometry (more working with ratios)

Ratios are found within a chemical equation.

2HCl + Ba(OH)2 2H2O + BaCl2 1 1

2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O and 1 mole of BaCl2

coefficients give MOLE RATIOS

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When N2O5 is heated, it decomposes:

2N2O5(g) 4NO2(g) + O2(g)

a. How many moles of NO2 can be produced from 4.3 moles of N2O5?

= moles NO2

4.3 mol N2O5

52

2

ON mol2

NO mol48.6

b. How many moles of O2 can be produced from 4.3 moles of N2O5?

= mole O2

4.3 mol N2O5

52

2

ON 2mol

O mol12.2

2N2O5(g) 4NO2(g) + O2(g)4.3 mol ? mol

2N2O5(g) 4NO2(g) + O2(g)4.3 mol ? mol

Mole – Mole Conversions

Units match

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When N2O5 is heated, it decomposes:2N2O5(g) 4NO2(g) + O2(g)

a. How many moles of N2O5 were used if 210g of NO2 were produced?

= moles N2O5

210 g NO2

2

52

NO mol4

ON mol22.28

b. How many grams of N2O5 are needed to produce 75.0 grams of O2?

= grams N2O5

75.0 g O2

2

52

O 1mol

ON mol2506

2

2

NO g0.46

NO mol

2

2

O g 32.0

O mol

52

52

ON mol

ON g108

gram ↔ mole and gram ↔ gram conversions

2N2O5(g) 4NO2(g) + O2(g)210g? moles

2N2O5(g) 4NO2(g) + O2(g)75.0 g? grams

Units match

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Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?

First write a balanced equation.

Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 3

Gram to Gram Conversions

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Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 3

Now let’s get organized. Write the information below the substances.

3.45 g ? grams

Gram to Gram Conversions

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Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 33.45 g ? grams

Let’s work the problem.

= g AlCl3

3.45 g Al

Alg 27.0

Almol

We must always convert to moles.

Now use the molar ratio.

Almol 2

AlClmol 2 3

Now use the molar mass to convert to grams.

3

3

AlClmol

AlClg 133.317.0

Units match

gram to gram conversions

Limiting Reagents/ Reactants• When compounds react, one compound will

disappear before the other compound.

• To determine which reactant will disappear first (this is called the limiting reagent) you will need to convert the reactants to mole or grams of one of the products (it must be the same product)

• When you convert, the answer that is the smallest is from the limiting reagent.

Limiting Reagents/Reactant

• What is the limiting reactant when 1 mol of hydrogen reacts with 1 mol of oxygen to produce water?

• First write the balanced chemical equation.• Convert 1 mole of each to moles of water. Will be

shown on board.• To determine how many moles of water are

produced, you use the answer that is the smallest. • In this case, hydrogen will run out before oxygen.

When hydrogen runs out the reaction stops only producing that amount of water.

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Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?

b. Determine the limiting reactant.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

First copy down the the BALANCED

equation!

Now place numerical the

information below the compounds.

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4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?

b. Determine the limiting reactant.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

0.15 mol 0.10 mol ? moles

Two starting amounts?

Where do we start?

Hide

one

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4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?b. Determine the limiting reactant.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

0.15 mol 0.10 mol ? molesHide

Based on:KO2 = mol O2

0.15 mol KO2

2

2

KO 4mol

O mol30.1125

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4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?b. Determine the limiting reactant.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)



0.15 mol KO2

2

2

KO 4mol

O mol30.1125

Hide

Based on: H2O

= mol O20.10 mol H2O

OH 2mol

O mol3

2

2 0.150


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4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?

Determine the limiting reactant.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)



0.15 mol KO2

2

2

KO 4mol

O mol30.1125

Based on: H2O

= mol O20.10 mol H2O

OH 2mol

O mol3

2

2 0.150

What is the theoretical yield? Hint: Which is the smallest

amount? What is the limiting reactant?

It was limited by theamount of KO2.

H2O = excess (XS) reactant!

Percent Yield

• Percent yield tells gives us a comparison of how much product you produce in lab compared to a maximum amount of product that could be produced.

• % yield = experimental value / calculated value x 100%

• You use stoichiometry to determine the calculated value.

Percent Yield

• In lab you react 2.00 g of hydrogen gas with 2.00 g of oxygen gas. You obtain 2.1 g of water vapor. What is your % yield?

• 1st you need to compute a limiting reagent problem to see how much water should be produced.

• 2nd you calculate % yield

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Theoretical yield vs. Actual yield

Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed, but only 12.3 grams of product were recovered. Determine the % yield.

Theoretical yield = 19.5 g based on limiting reactant

Actual yield = 12.3 g experimentally recovered

100x yield ltheoretica

yield actual yield %

yield 63.1% 100x 19.5

12.3 yield %

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4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced?

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

120.0 g 47.0 g ? gHide one

Based on:KO2

= g O2 120.0 g KO2

g1.71

mol2

2

KO 4mol

O mol3

2

2

O mol

O g0.3240.51

Limiting/Excess Reactant Problem with % Yield

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4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)


4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

120.0 g 47.0 g ? g

Based on:KO2

= g O2 120.0 g KO2

g1.71

mol2

2

KO 4mol

O mol3

2

2

O mol

O g0.3240.51

Based on:H2O

= g O2

Question if only 35.2 g of O2 were recovered, what was the percent yield?

yield 86.9% 100x 51.40

2.35 100x

ltheoretica

actual

Hide

47.0 g H2O

OH g 02.18

OH mol

2

2

OH mol 2

O mol 3

2

2

2

2

O mol

O g0.32125.3

Limiting/Excess Reactant Problem with % Yield

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4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

120.0 g 47.0 g ? g

Based on:KO2

= g O2 120.0 g KO2

g1.71

mol2

2

KO 4mol

O mol3

2

2

O mol

O g0.3240.51

Based on:H2O

= g O247.0 g H2O

OH g 02.18

OH mol

2

2

OH mol 2

O mol 3

2

2

2

2

O mol

O g0.32125.3

Determine how many grams of Water were left over.The Difference between the above amounts is directly RELATED to the excess H2O.

125.3 - 40.51 = 84.79 g of O2 that could have been formed from the excess water.

= g XS H2O84.79 g O2

2

2

O g 32.0

O mol

2

2

O mol 3

OH mol 2

OH mol 1

OH g 02.18

2

2 31.83

Application of topics

• Stoichiometry is used to determine maximum yield when performing any type of reaction.

• You will use Stoichiometry, Limiting Reagents, Percent Yield, and Percent Error when performing labs.

• It is necessary to do these calculations to determine your accuracy in lab. I will be looking at your accuracy when I grade labs.

stoichiometry tutorial

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