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STOICHIOMETRY

Stoichiometry

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Page 1: Stoichiometry

STOICHIOMETRY

Page 2: Stoichiometry

What is stoichiometry?

Stoichiometry is the quantitative study of reactants and products in a chemical reaction.

Page 3: Stoichiometry

What You Should Expect

Given : Amount of reactants Question: how much of products can be

formed. Example

2 A + 2B 3C Given 20.0 grams of A and sufficient B,

how many grams of C can be produced?

Page 4: Stoichiometry

What do you need?

You will need to use

i. molar ratios,

ii. molar masses,

iii. balancing and interpreting equations, and

iv. conversions between grams and moles.

Note: This type of problem is often called "mass-mass."

Page 5: Stoichiometry

Steps Involved in Solving Mass-Mass Stoichiometry Problems

Balance the chemical equation correctly Using the molar mass of the given substance,

convert the mass given to moles. Construct a molar proportion (two molar

ratios set equal to each other) Using the molar mass of the unknown

substance, convert the moles just calculated to mass.

Page 6: Stoichiometry

Mole Ratios

A mole ratio converts moles of one compound in a balanced chemical equation into moles of another compound.

Page 7: Stoichiometry

Example

Reaction between magnesium and oxygen to form magnesium oxide. ( fireworks)

2 Mg(s) + O2(g) 2 MgO(s)

Mole Ratios:

2 : 1 : 2

Page 8: Stoichiometry

1) N2 + 3 H2 ---> 2 NH3

Write the mole ratios for N2 to H2 and NH3 to H2.

2) A can of butane lighter fluid contains 1.20 moles of butane (C4H10). Calculate the number of moles of carbon dioxide given off when this butane is burned.

Practice Problems

Page 9: Stoichiometry

Mole-Mole Problems Using the practice question 2) above: Equation of reaction

2C4H10 + 13O2 8CO2 + 10H2O Mole ratio

C4H10 CO2

1 : 4 [ bases] 1.2 : X [ problem]

By cross-multiplication, X = 4.8 mols of CO2 given off

Page 10: Stoichiometry

Mole-Mass Problems

Problem 1: 1.50 mol of KClO3 decomposes. How many grams of O2 will be produced? [k = 39, Cl = 35.5, O = 16]

2 KClO3 2 KCl + 3 O2

Page 11: Stoichiometry

Three steps…Get Your Correct Answer

Use mole ratioGet the answer in moles and thenConvert to Mass. [Simple

Arithmetic]

Hello!If you are given a mass in the

problem, you will need to convert this to moles first. Ok?

Page 12: Stoichiometry

Let’s go!2 KClO3 2 KCl + 3 O2

2 : 31.50 : XX = 2.25molConvert to mass

2.25 mol x 32.0 g/mol = 72.0 grams

Cool!

Page 13: Stoichiometry

Try This: We want to produce 2.75 mol of KCl. How many

grams of KClO3 would be required?

Soln

KClO3 : KCl

2 : 2X : 2.75X = 2.75mol

In mass: 2.75mol X 122.55 g/mol

= 337 grams zooo zimple!

Page 14: Stoichiometry

Mass-Mass Problems

There are four steps involved in solving these problems:

Make sure you are working with a properly balanced equation.

Convert grams of the substance given in the problem to moles.

Construct two ratios - one from the problem and one from the equation and set them equal. Solve for "x," which is usually found in the ratio from the problem.

Convert moles of the substance just solved for into grams.

Page 15: Stoichiometry

Just follow mass-mass problem to the penultimate level

Mass-Volume Problems

Page 16: Stoichiometry

Like this:There are four steps involved in solving

these problems: Make sure you are working with a

properly balanced equation. Convert grams of the substance given

in the problem to moles. Construct two ratios - one from the

problem and one from the equation and set them equal. Solve for "x," which is usually found in the ratio from the problem.

Convert moles of the substance just solved for into Volume.

Page 17: Stoichiometry

Conversion of mole to volume

No of moles = VolumeMolar volume

Can you remember a similar equation?

Page 18: Stoichiometry

Molar volume

The molar volume is the volume occupied by one mole of ideal gas at STP. Its value is: 22.4dm3

Page 19: Stoichiometry

Practice Problems

Calculate the volume of carbon dioxide formed at STP in ‘dm3' by the complete thermal decomposition of 3.125 g of pure calcium carbonate (Relative atomic mass of Ca=40, C=12, O=16)

Solution: Convert the mass to mole:Molar mass of CaCO3

= 40 + 12 + (16 x 3) = 100gmol-1

Mole = mass/molar mass3.125/100 = 0.03125mol

Page 20: Stoichiometry

Practice Problems

As per the equation,

Mole ratio 1 : 1problem 0.03125mol X

X = 0.03125mol of CO2

Convert mole to volume [slide 17]

Volume = (0.03125 x 22.4)dm3

= 0.7dm3