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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
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.
- What reacts? Metane (CH4) and oxygen gas (O2)
- 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.