The equation `2Al(s)+ 3/2O_2 to Al_2O_3(s)` shows that

Options

A. 2 moles of aluminium react with 3/2 moles of oxygen to produce one mole of aluminium oxide
B. 2 atoms of aluminium react with 3/2 atoms of oxygen to produce one atom of aluminium oxide
C. 2g of aluminium react with 3/2 g of oxygen to produce 1 g of aluminium oxide
D. 2g of aluminium react with 3/2 litres of oxygen to produce 1 g of aluminium oxide

Answer

To solve the question regarding the reaction \(2Al(s) + \frac{3}{2}O_2 \rightarrow Al_2O_3(s)\), we will analyze the stoichiometry of the reaction step by step.

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- The reactants are aluminum (Al) and oxygen (O₂).
- The product is aluminum oxide (Al₂O₃).

2. **Write the Balanced Chemical Equation**:
- The given equation is already balanced:
\[ 2Al(s) + \frac{3}{2}O_2 \rightarrow Al_2O_3(s) \]

3. **Determine the Molar Ratios**:
- From the balanced equation, we can see that:
- 2 moles of aluminum react with \(\frac{3}{2}\) moles of oxygen to produce 1 mole of aluminum oxide.
- This means:
- For every 2 moles of Al, \(\frac{3}{2}\) moles of O₂ are required, resulting in 1 mole of Al₂O₃.

4. **Understanding Stoichiometry**:
- The coefficients in the balanced equation represent the stoichiometric coefficients, which indicate the proportions of reactants and products involved in the reaction.
- In this case, the stoichiometric coefficients are:
- 2 for Al
- \(\frac{3}{2}\) for O₂
- 1 for Al₂O₃

5. **Conclusion**:
- The equation shows that 2 moles of aluminum react with 1.5 moles of oxygen to produce 1 mole of aluminum oxide. This is a stoichiometric relationship that can be used for calculations involving moles, mass, and volume in chemical reactions.