STOICHIOMETRY

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

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."

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

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

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

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?

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!

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!

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.

Just follow mass-mass problem to the penultimate level

Mass-Volume Problems

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 = VolumeMolar volume

Can you remember a similar equation?

Molar volume

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 mass3.125/100 = 0.03125mol

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

This powerpoint was kindly donated to www.worldofteaching.com

http://www.worldofteaching.com is home to over a thousand powerpoints submitted by teachers. This is a completely free site and requires no registration. Please visit and I hope it will help in your teaching.