To determine which of the statements are correct, we will analyze each statement step by step.
### Step 1: Analyze the first statement
**Statement i:** 21.0 g of lithium reacts with 32.0 g of O₂.
**Reaction:**
\[ 4 \text{Li} + \text{O}_2 \rightarrow 2 \text{Li}_2\text{O} \]
**Calculating moles of lithium (Li):**
- Molar mass of Li = 7 g/mol
- Given mass of Li = 21 g
\[
\text{Moles of Li} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{21 \, \text{g}}{7 \, \text{g/mol}} = 3 \, \text{moles}
\]
**Calculating moles of oxygen (O₂):**
- Molar mass of O₂ = 32 g/mol
- Given mass of O₂ = 32 g
\[
\text{Moles of O}_2 = \frac{32 \, \text{g}}{32 \, \text{g/mol}} = 1 \, \text{mole}
\]
**Determining the limiting reagent:**
From the balanced equation, 4 moles of Li react with 1 mole of O₂. Therefore, for 3 moles of Li, the required moles of O₂ can be calculated as follows:
\[
\text{Required moles of O}_2 = \frac{3 \, \text{moles of Li}}{4} = 0.75 \, \text{moles of O}_2
\]
Since we have 1 mole of O₂ available, it is in excess. Thus, Li is the limiting reagent.
### Step 2: Analyze the second statement
**Statement ii:** 3.9 g of K reacts with 4.26 g of Cl₂.
**Reaction:**
\[ 2 \text{K} + \text{Cl}_2 \rightarrow 2 \text{KCl} \]
**Calculating moles of potassium (K):**
- Molar mass of K = 39 g/mol
- Given mass of K = 3.9 g
\[
\text{Moles of K} = \frac{3.9 \, \text{g}}{39 \, \text{g/mol}} = 0.1 \, \text{moles}
\]
**Calculating moles of chlorine (Cl₂):**
- Molar mass of Cl₂ = 71 g/mol
- Given mass of Cl₂ = 4.26 g
\[
\text{Moles of Cl}_2 = \frac{4.26 \, \text{g}}{71 \, \text{g/mol}} \approx 0.06 \, \text{moles}
\]
**Determining the limiting reagent:**
From the balanced equation, 2 moles of K react with 1 mole of Cl₂. Therefore, for 0.1 moles of K, the required moles of Cl₂ can be calculated as follows:
\[
\text{Required moles of Cl}_2 = \frac{0.1 \, \text{moles of K}}{2} = 0.05 \, \text{moles of Cl}_2
\]
Since we have 0.06 moles of Cl₂ available, it is in excess. Thus, K is the limiting reagent.
### Step 3: Calculate the products formed
**For the first reaction:**
- Moles of Li₂O formed from 3 moles of Li:
\[
\text{Moles of Li}_2\text{O} = \frac{3 \, \text{moles of Li}}{2} = 1.5 \, \text{moles of Li}_2\text{O}
\]
- Mass of Li₂O formed:
\[
\text{Mass of Li}_2\text{O} = 1.5 \, \text{moles} \times 30 \, \text{g/mol} = 45 \, \text{g}
\]
**For the second reaction:**
- Moles of KCl formed from 0.1 moles of K:
\[
\text{Moles of KCl} = 0.1 \, \text{moles of K} \rightarrow 0.1 \, \text{moles of KCl}
\]
- Mass of KCl formed:
\[
\text{Mass of KCl} = 0.1 \, \text{moles} \times 74.5 \, \text{g/mol} = 7.45 \, \text{g}
\]
### Conclusion
- **Statement i:** O₂ is in excess, and Li is the limiting reagent. **True.**
- **Statement ii:** K is the limiting reagent, and Cl₂ is in excess. **True.**
- Mass of Li₂O formed is 45 g. **True.**
- Mass of KCl formed is 7.45 g. **True.**
All statements are correct.
