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STOICHIOMETRY – Chapter 11 PowerPoint PPT Presentation

STOICHIOMETRY – Chapter 11. What is stoichiometry?. Stoichiometry is the quantitative study of reactants and products in a chemical reaction. What You Should Expect. Given : Amount of reactants Question: how much of products can be formed. Example 2 A + 2B 3C

STOICHIOMETRY – Chapter 11

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STOICHIOMETRY – Chapter 11

What is stoichiometry?

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

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?

What do you need?

You will need to use

• molar ratios,

• molar masses,

• balancing and interpreting equations, and

• conversions between grams and moles.

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

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.

Mole Ratios

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

Example

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

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

Mole Ratios:

2 : 1: 2

Practice Problems

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.

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

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

• Use mole ratio

• Get the answer in moles and then

• Convert to Mass. [Simple Arithmetic]

Hello!

If you are given a mass in the problem, you will need to convert this to moles first. Ok?

Let’s go!

2 KClO3 2 KCl + 3 O2

2: 3

1.50 : X

X = 2.25mol

Convert to mass

2.25 mol x 32.0 g/mol = 72.0 grams

Cool!

Try This:

• We want to produce 2.75 mol of KCl. How many grams of KClO3 would be required?

Soln

KClO3 :KCl

2 : 2

X :2.75

X = 2.75mol

In mass: 2.75mol X 122.55 g/mol

= 337 grams zooo zimple!

Mass-MassProblems

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.

Mass-VolumeProblems

Just follow mass-mass problem to the penultimate level

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.

Conversion of mole to volume

No of moles = Volume

Molar volume

Can you remember a similar equation?

Molarvolume

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

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 mass

3.125/100 = 0.03125mol

Practice Problems

As per the equation,

Mole ratio1 : 1

problem0.03125mol X

X = 0.03125mol of CO2

Convert mole to volume [slide 17]

Volume = (0.03125 x 22.4)dm3

=0.7dm3