To solve the problem, we need to identify the element X, the compound Y formed when X reacts with hydrogen, and the compound Z formed when X reacts with chlorine.
### Step-by-Step Solution:
1. **Identify Element X**:
- Element X is from Group 2 of the periodic table, which includes elements like Beryllium (Be), Magnesium (Mg), Calcium (Ca), Strontium (Sr), and Barium (Ba). These elements typically form +2 cations (X²⁺).
- Given the context of the problem, we will consider Calcium (Ca) as a potential candidate for X.
2. **Reaction with Hydrogen**:
- When X reacts with hydrogen gas (H₂) at 200°C, it forms a compound Y. The reaction can be represented as:
\[
\text{X} + \text{H}_2 \rightarrow \text{Y}
\]
- For calcium, this reaction would yield Calcium Hydride (CaH₂):
\[
\text{Ca} + \text{H}_2 \rightarrow \text{CaH}_2
\]
- Thus, compound Y is Calcium Hydride (CaH₂).
3. **Decomposition of Compound Y**:
- When compound Y (CaH₂) is heated, it decomposes back into element X and hydrogen gas:
\[
\text{CaH}_2 \rightarrow \text{Ca} + \text{H}_2
\]
- The problem states that for 1.00 g of X reacted, 559 ml of H₂ is produced at STP.
4. **Calculating Molar Ratios**:
- At STP, 1 mole of gas occupies 22400 ml. Therefore, we can calculate the moles of H₂ produced:
\[
\text{Moles of H}_2 = \frac{559 \text{ ml}}{22400 \text{ ml/mol}} \approx 0.0249 \text{ mol}
\]
- Since 1 mole of Ca produces 1 mole of H₂, the moles of Ca that reacted is also 0.0249 mol.
- The molar mass of Ca is approximately 40 g/mol. Thus, the mass of Ca that corresponds to 0.0249 mol is:
\[
\text{Mass of Ca} = 0.0249 \text{ mol} \times 40 \text{ g/mol} \approx 1.00 \text{ g}
\]
- This confirms that X is indeed Calcium (Ca).
5. **Identify Compound Z**:
- Now, we need to determine compound Z formed when X (Ca) reacts with chlorine (Cl₂).
- The reaction is:
\[
\text{Ca} + \text{Cl}_2 \rightarrow \text{CaCl}_2
\]
- To confirm the identity of Z, we need to check the percentage by mass of chlorine in CaCl₂.
- The molar mass of CaCl₂ is:
\[
\text{Molar mass of CaCl}_2 = 40 \text{ (Ca)} + 2 \times 35.5 \text{ (Cl)} = 111 \text{ g/mol}
\]
- The mass percentage of chlorine in CaCl₂ is:
\[
\text{Percentage of Cl} = \left( \frac{2 \times 35.5}{111} \right) \times 100 \approx 63.89\%
\]
- This matches the given percentage in the problem, confirming that compound Z is Calcium Chloride (CaCl₂).
### Final Identification:
- **Element X**: Calcium (Ca)
- **Compound Y**: Calcium Hydride (CaH₂)
- **Compound Z**: Calcium Chloride (CaCl₂)
