To solve the problem, we need to balance the given chemical equations and find the values of \( x \), \( y \), and \( z \). Then, we will sum these values.
### Step 1: Balancing the first equation
The first equation is:
\[ \text{Mg} + x \text{HNO}_3 \rightarrow \text{Mg(NO}_3\text{)}_2 + \text{H}_2 \]
1. **Identify the products**:
- Magnesium (Mg) is balanced on both sides (1 Mg).
- In the product side, we have 2 nitrogen atoms (from \(\text{Mg(NO}_3\text{)}_2\)), which means we need 2 nitrogen atoms on the reactant side. Therefore, \( x \) must be 2 because each \(\text{HNO}_3\) contributes 1 nitrogen atom.
- For oxygen, we have \( 2 \times 3 = 6 \) oxygen atoms on the product side, and with \( x = 2 \), we also have \( 2 \times 3 = 6 \) oxygen atoms on the reactant side.
- Hydrogen: We have 2 hydrogen atoms from \( 2 \text{HNO}_3\) which matches the 2 hydrogen atoms from \( \text{H}_2\).
Thus, \( x = 2 \).
### Step 2: Balancing the second equation
The second equation is:
\[ \text{Cu} + y \text{HNO}_3 \rightarrow 3 \text{Cu(NO}_3\text{)}_2 + 2 \text{NO} + 4 \text{H}_2\text{O} \]
1. **Identify the products**:
- Copper (Cu): There are 3 Cu on the product side, so we need 3 Cu on the reactant side.
- Nitrogen: From \( 3 \text{Cu(NO}_3\text{)}_2\), we have \( 3 \times 2 = 6 \) nitrogen atoms, and from \( 2 \text{NO}\), we have 2 nitrogen atoms, totaling 8 nitrogen atoms on the product side. Therefore, \( y \) must be 8 because \( y \) HNO3 contributes \( y \) nitrogen atoms.
- Oxygen: From \( 3 \text{Cu(NO}_3\text{)}_2\), we have \( 3 \times 6 = 18 \) oxygen atoms, from \( 2 \text{NO}\) we have 2, and from \( 4 \text{H}_2\text{O}\) we have 4, totaling 24 oxygen atoms. With \( y = 8\), we have \( 8 \times 3 = 24\) oxygen atoms on the reactant side.
- Hydrogen: We have \( 8 \) hydrogen atoms from \( 8 \text{HNO}_3\) which matches \( 4 \text{H}_2\text{O}\) (4 H).
Thus, \( y = 8 \).
### Step 3: Balancing the third equation
The third equation is:
\[ \text{I}_2 + z \text{HNO}_3 \rightarrow 2 \text{HIO}_3 + 10 \text{NO}_2 + 4 \text{H}_2\text{O} \]
1. **Identify the products**:
- Iodine: There are 2 iodine atoms on the product side, so we need 2 iodine atoms on the reactant side.
- Nitrogen: From \( 10 \text{NO}_2\), we have 10 nitrogen atoms. Therefore, \( z \) must be 10 because \( z \) HNO3 contributes \( z \) nitrogen atoms.
- Oxygen: From \( 2 \text{HIO}_3\), we have \( 2 \times 3 = 6\) oxygen atoms, from \( 10 \text{NO}_2\) we have \( 10 \times 2 = 20\), and from \( 4 \text{H}_2\text{O}\) we have 4, totaling 30 oxygen atoms. With \( z = 10\), we have \( 10 \times 3 = 30\) oxygen atoms on the reactant side.
- Hydrogen: We have \( 2 \) from \( 2 \text{HIO}_3\) and \( 4 \) from \( 4 \text{H}_2\text{O}\), totaling 10 hydrogen atoms. With \( z = 10\), we have \( 10 \) hydrogen atoms on the reactant side.
Thus, \( z = 10 \).
### Step 4: Calculate the sum of \( x \), \( y \), and \( z \)
Now we can find the sum:
\[ x + y + z = 2 + 8 + 10 = 20 \]
### Final Answer
The sum of \( x \), \( y \), and \( z \) is \( 20 \).
