...means the mole ratio in a reaction.

So... 2H₂ + O₂--> 2H₂O

....tells us that it takes 2 Hydrogen molecules to react with one Oxygen molecule, making 2 water molecules.

In reality, it tells us that it takes 2 moles of Hydrogen to react with one mole of Oxygen, making 2 moles of water.

But when we mix Hydrogen and Oxygen it is likely that either

a) The Hydrogen runs out leaving some Oxygen or

b) The Oxygen runs out leaving some Hydrogen

Whichever reactant is left over is said to be in excess.

Whichever runs out is the limiting reactant.

How do we know which is which?

C₆H₁₂O₆+6O₂→6CO₂+6H₂O

What mass of Carbon dioxide forms in the reaction of 25 grams of Glucose with 40 grams of Oxygen?

First we need to know how many moles of each reactant there are.

C₆H₁₂O₆ : 25g/180.06g/mol=0.1388mol

O₂: 40g/32g/mol=1.25mol

The equation tells us we need 6 moles of O₂ for every 1 mole of Glucose.

6 x 0.1388 = 0.8328

So, we have more O₂ then needed – it is in excess.

Glucose is the limiting reagent – we use this for the calculation.

The equation tells us we make 6 moles of CO₂ for every 1 mole of Glucose

So, we make 0,8328 moles of Carbon Dioxide

1. Calculate the mass of magnesium oxide possible if 2.40 g of Mg reacts with 10.0 g of O₂

2Mg+O₂→2MgO

2. What is the limiting reagent if 76.4 grams of C₂H₃Br₃ were reacted with 49.1 grams of O₂? How much water is made?

C₂H₃Br₃+11O₂→8CO₂+6H₂O+6Br₂

3. What is the limiting reagent if 78 grams of Na₂O were reacted with 29.4 grams of H₂O? How much NaOH is made?

Na₂O + H₂O → 2NaOH

4. How much the excess reagent remains if 24.5 grams of CoO is reacted with 2.58 grams of O₂?

4CoO+O₂→2Co₂O₃

The theoretical yield is what you expect to make.

N₂ + 3H₂ → 2NH₃

If we start with 56g of Nitrogen and make 60g of Ammonia, what is the percentage yield?

Experimental Yield = 60g – it tells us this in the question.

Theoretical yield = We are starting with 56g of N₂ which is 56/28 = 2 moles

We should be able to make 4 moles of Ammonia

4 moles of ammonia = 4 x 17 = 68g --- this is the theoretical yield.

Percentage yield = (actual yield/theoretical yield) x 100

Percentage Yield = 60/ 68 x 100 = 88.2 %

· LEARN – the assumptions of the Ideal Gases:

i) Gas particles are so small and the space between them is so big that effectively GAS PARTICLES OCCUPY NO SPACE

ii) GAS PARTICLES MOVE QUICKLY & FREQUENTLY CHANGE DIRECTION

iii) GAS PARTICLES ARE NOT ATTRACTED TO EACH OTHERS

iv) ALL COLLISIONS BETWEEN PARTICLES ARE ELASTIC (no energy is lost)

Avogadro never even attempted to calculate Avogadro’s Constant – he said that equal volumes of any gas occupy the same space (See point 1)

We now say that 1 mole of any gas occupies 22.7 dm³ at Standard Temp. and Pressure (STP = 0^oC & 100kPa)

Gay-Lussac said reacting volumes of gases could always be represented as whole number ratios.

Between the two we had the beginning of mole calculations.

If 13.46 dm³ of N₂ reacts at STP what volume of Hydrogen is required, what volume of Ammonia is made? How many moles of Ammonia is this?

N₂ + 3H₂ --> 2NH₃

· Volume H₂ = 3 x volume N₂ = 3 x 13.46 = 40.38 dm³

· Volume NH₃ = 2 x volume N₂ = 2 x 13.46 = 26.92 dm³

· Moles (N₂) = 13.46/22.7 = 0.598

· Moles NH₃ = 2 x moles N₂ = 2 x 0.598= 1.196 mol

1. If 15.2dm³ of CH₄ reacts at STP what volume of Oxygen is required, what volume of CO₂ is made? How many moles of CO₂ is this?

2O₂ + CH₄ --> CO₂ + 2H₂O

2. If 5.1dm³ of C₃H₈ reacts at STP what volume of Oxygen is required, what volume of CO₂ is made? How many moles of CO₂ is this?

5O₂ + C₃H₈ --> 3CO₂ + 4H₂O

1 atm = 100,000Pa

1. Calculate the number of moles of Oxygen in 2.6 dm3 at 1.01 x 10⁵ Pa and 300K

2 Given the following sets of values, calculate the unknown quantity.

a) P = 1.01 atm V = ? n = 0.00831 mol T = 25°C

b) P = ? V= 0.602 L n = 0.00801 mol T = 311 K

3. 3) At what temperature would 2.10 moles of N₂ gas have a pressure of 1.25 atm and in a 25.0 dm³ tank?

4) What volume is occupied by 5.03 g of O₂ at 28°C & a pressure of 0.998atm?

5) Find the pressure in a 212 dm³ tank containing 23.3 kg of argon at 25°C?

1 What is the concentration of Calcium Chloride if 18.2g is dissolved in 400cm³?

M_r (CaCl₂) = 20 + 35.5 + 35.5 = 91 g/mol

Moles (CaCl₂) = Mass/M_r =18.2/91 = 0.2 mol

Concentration = Moles/ Vol (dm³) = 0.2/(400/1000) = 05 mol/dm³

2. What is the concentration of Sodium Chloride (NaCl) if 8.7g is dissolved in 500cm³?

3. What is the concentration of Aluminium Chloride (AlCl₃) if 26.2g is dissolved in 800cm³?

4. What is the concentration of Copper Sulphate (CuSO₄. 5H₂O) if 23.6g is dissolved in 500cm³?

5. How many grams of NaCl are contained in a 250 cm³ of a solution with concentration 0.4 mol/dm³?

6. How many grams of CaCl₂ are contained in a 250 cm³ of a solution with concentration 0.4 mol/dm³?

7. A solution of 0.65 mol/dm³ NaCl contains 2.3g. What volume of water was the NaCl dissolved in?

8. A solution of 0.33 mol/dm³ CaCl₂ contains 4.5g. What volume of water was the CaCl₂ dissolved in?

Solutions are often made up in concentrations of grams per dm³.

These are not very useful for mole calculations, so we usually convert them to mol/dm³ anyway.

Moldm^-3 = gdm^-3/M_r

1 What is the molarity (in mol/dm³) of a solution of NaOH of concentration 2.4 g/dm³?

2 What is the molarity (in mol/dm³) of a solution of Ca(OH)₂ of concentration 1.4 g/dm³?

3 What is the molarity (in mol/dm³) of a solution of NaSO₄ of concentration 0.4 g/dm³?

4 A solution of Ca(HCO₃)₂ has a concentration of 0.45 mol/dm³. What is its concentration in g/dm³?

5 A solution of Mg(HSO₄)₂ has a concentration of 0.15 mol/dm³. What is its concentration in g/dm³?

1. Find the concentration of a solution of 15 cm³ Ca(OH)₂ that required 23.5 cm³ of a solution HCl (0.75mol/dm³) to neutralise.

Ca(OH)₂ + 2HCl --> CaCl₂ + 2H₂O

2. Find the concentration of a solution of 35 cm³ NaOH that required 13.5 cm³ of a solution HCl (0.5mol/dm³) to neutralise.

NaOH + HCl--> NaCl + H₂O

3. Find the concentration of a solution of 35 cm³ Ca(OH)₂ that required 12.5 cm³ of a solution H₂SO₄ (0.5mol/dm³) to neutralise.

Ca(OH)₂ + H₂SO₄--> Ca SO₄ + 2H₂O