Hydrogen reacts with oxygen producing water.

Word equation:

• Hydrogen gas + oxygen gas → water

Chemical equation

• 2H2 + O2 → 2H2O

Why is the chemical reaction balanced? Stoichiometry

Stoichiometry = relative quantities in chemical reactions

Two hydrogen molecules react with one oxygen molecule to produce two water molecules.

Everything on the left side of the reaction arrow turns into products on the right side of the reaction arrow. The number of atoms on the left is equal to the number of atoms on the right ⇒ the mass is conserved.

Conservation of mass: No atoms disappear or appear during the reaction.

What does the chemical equation tell us?

2H2 + O2 → 2H2O

Two hydrogen molecules and one oxygen molecule are needed to produce two water molecules.

The amount of hydrogen molecules required to make water is double the amount of oxygen molecules.

Relative quantities: $$n_{\text{H}_2} : n_{\text{O}_2} : n_{\text{H}_2\text{O}} = 2:1:2$$

These relative quantities are always valid for this reaction. Examples:

 2H2 + O2 → 2H2O 2 molecules 1 molecule 2 molecules 2 pairs 1 pair 2 pairs 2 dozens 1 dozen 2 dozens 4 dozens 2 dozens 4 dozens 2 mol 1 mol 2 mol 1 mol 0.5 mol 1 mol 2.50 mol 1.25 mol 2.50 mol

The mass is conserved in the reaction

The mass of the reactants is exactly equal to the mass of the products. An example:

 2H2 + O2 → 2H2O $1 \text{mol}$ $0.5\text{mol}$ → $1 \text{mol}$ $m = Mn$ $2.016 \text{g/mol} \times 1 \text{mol} =\\= 2.016 \text{g}$ $32.0 \text{g/mol} \times 0.5 \text{mol} =\\= 16.0 \text{g}$ → $18.016 \text{g/mol} \times 1 \text{mol} =\\= 18.016 \text{g}$ $2.016 \text{g}$ $16.0 \text{g}$ → $18.016 \text{g}$

The mass to the left of the reaction arrow is equal to the mass to the right:

$2.016 \text{g} + 16.0 \text{g} = 18.016 \text{g}$

Combustion of methane

What happens when methane, CH4, is combusted?

We have to know a few things.

1. What reacts? Metane (CH4) and oxygen gas (O2)
2. What is produced? Carbon dioxide (CO2) and water (H2O)
• When hydrocarbons and carbohydrates are combusted, carbon dioxide and water is always produced.

Word equation:

• Methane + oxygen gas → carbon dioxide + water

Chemical equation is balanced to:

• CH4 + 2O2 → CO2 + 2H2O

Relative quantities: $${n_{{\text{C}}{{\text{H}}_4}}}:{n_{{{\text{O}}_2}}}:{n_{{\text{C}}{{\text{O}}_2}}}:{n_{{{\text{H}}_2}{\text{O}}}} = 1:2:1:2$$

These relative quantities are always valid for this reaction. Examples:

 CH4 + 2O2 → CO2 + 2H2O 1 molecule 2 molecules 1 molecule 2 molecules 1 pair 2 pairs 1 pair 2 pairs 1 dozen 2 dozens 1 dozen 2 dozens 1 mol 2 mol 1 mol 2 mol 3.45 mol 6.90 mol 3.45 mol 6.90 mol

The mass is conserved in the reaction – as always!

The mass of the reactants is exactly equal to the mass of the products. An example:

 CH4 + 2O2 → CO2 + 2H2O $1 \text{mol}$ $2 \text{mol}$ → $1 \text{mol}$ $2 \text{mol}$ $m = Mn$ $16 \text{g/mol} \times 1 \text{mol} =\\= 16 \text{g}$ $32 \text{g/mol} \times 2 \text{mol} =\\= 64 \text{g}$ → $44 \text{g/mol} \times 1 \text{mol} =\\= 44 \text{g}$ $18 \text{g/mol} \times 2 \text{mol} =\\= 36 \text{g}$ $16 \text{g}$ $64 \text{g}$ → $44 \text{g}$ $36 \text{g}$

The mass to the left of the reaction arrow is equal to the mass to the right:

$\underbrace {16{\text{g}} + 64{\text{g}}}_{ = 80{\text{g}}} = \underbrace {44{\text{g}} + 36{\text{g}}}_{ = 80{\text{g}}}$

Conservation of mass: No atoms disappear or appear during the reaction.