Summary of "Hukum Avogadro | Hipotesis Avogadro | Kimia Kelas 10"

Main ideas & concepts (Avogadro’s Law)

Avogadro’s law (hypothesis): At the same temperature and pressure, all gases with the same volume contain the same number of molecules.
Key proportionality:
- The ratio of gas volumes = the ratio of the number of molecules.
- Because coefficients in a balanced chemical equation represent mole ratios, it follows that:
  - Volume ratio = molecule ratio = coefficient ratio

If gases are at the same temperature and pressure, then for two species “1” and “2”: [ \frac{V_1}{V_2}=\frac{n_1}{n_2}=\frac{a_1}{a_2} ] where:

Write the balanced reaction equation for the given reactants → products.
Identify the known quantity (number of molecules of one reactant).
Use proportionality based on coefficients: [ \frac{n_{\text{reactant 1}}}{a_{\text{reactant 1}}} = \frac{n_{\text{product}}}{a_{\text{product}}} ] Equivalently, molecule ratio = coefficient ratio.
Multiply/divide using the coefficient ratio to get the number of molecules of the requested substance.

Write the general reaction form (e.g., ( \text{Cl}_2 + \text{O}_2 \rightarrow \text{Cl}_x\text{O}_y )).
Use the known volumes to form a ratio, simplifying into a smallest integer ratio.
Assign that ratio to the stoichiometric coefficients (the simplified volume ratio matches the balanced coefficient ratio when (T) and (P) are the same).
Balance atoms to determine (x) and (y):
- Equate the number of each type of atom on both sides.
- Solve for (x) and (y) from those equalities.

Balance the reaction first to get the coefficient ratio.
Compare reactant usage by computing:
- ( \frac{V}{\text{coefficient}} ) for each reactant.
The smaller value corresponds to the reactant that is used up first (the limiting reactant).
Use the coefficient ratio to compute:
- Volume of product formed (from the limiting reactant’s consumed amount).
- Leftover reactant volume = initial volume − consumed volume.

Example 1 (molecules → product molecules)
- Reaction: ( \text{N}_2 + \text{H}_2 \rightarrow \text{NH}_3) (balanced using coefficients).
- Given: number of ( \text{H}_2) molecules = (7.5\times 10^{23}).
- Find: number of ( \text{NH}_3) molecules using the coefficient/molecule ratio.
Example 2 (volume → molecules)
- Given volumes of ( \text{N}_2) and asked molecules in ( \text{O}_2).
- Uses: volume ratio = molecule ratio at the same (T) and (P).
Example 3 (find (x) and (y) in ( \text{Cl}_x\text{O}_y))
- Given volumes: (60\text{ mL Cl}_2), (150\text{ mL O}_2), produces (60\text{ mL Cl_xO_y}).
- Uses simplified volume ratio → coefficient ratio → atom balancing.
- Solves: (x = 2) and (y = 5).
Example 4 (product volume and leftover volume via limiting reactant)
- Given: (10\text{ L SO}_2) and (6\text{ L O}_2).
- Reaction: ( \text{SO}_2 + \text{O}_2 \rightarrow \text{SO}_3) (balanced to get coefficients).
- (A) Find volume of ( \text{SO}_3) formed using limiting reactant determination.
- (B) Find leftover volume of the reactant in excess.