In the following equations
`Na_(2) CO_(3) + x HC1 to 2NaC1 + CO_(2) + H_(2) O`
the value of x is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the equation \( \text{Na}_2\text{CO}_3 + x \text{HCl} \rightarrow 2 \text{NaCl} + \text{CO}_2 + \text{H}_2\text{O} \) and find the value of \( x \), we need to balance the chemical equation. Here's a step-by-step solution:

### Step 1: Write down the unbalanced equation
The unbalanced equation is:
\[ \text{Na}_2\text{CO}_3 + x \text{HCl} \rightarrow 2 \text{NaCl} + \text{CO}_2 + \text{H}_2\text{O} \]

### Step 2: Count the number of atoms of each element on both sides
- **Reactants:**
- Sodium (Na): 2 (from \( \text{Na}_2\text{CO}_3 \))
- Carbon (C): 1 (from \( \text{Na}_2\text{CO}_3 \))
- Oxygen (O): 3 (from \( \text{Na}_2\text{CO}_3 \))
- Hydrogen (H): \( x \) (from \( x \text{HCl} \))
- Chlorine (Cl): \( x \) (from \( x \text{HCl} \))

- **Products:**
- Sodium (Na): 2 (from \( 2 \text{NaCl} \))
- Carbon (C): 1 (from \( \text{CO}_2 \))
- Oxygen (O): 3 (2 from \( \text{CO}_2 \) and 1 from \( \text{H}_2\text{O} \))
- Hydrogen (H): 2 (from \( \text{H}_2\text{O} \))
- Chlorine (Cl): 2 (from \( 2 \text{NaCl} \))

### Step 3: Set up equations based on the number of atoms
From the products, we can see that:
- Sodium: 2 on both sides (balanced)
- Carbon: 1 on both sides (balanced)
- Oxygen: 3 on both sides (balanced)
- Chlorine: \( x = 2 \) (since there are 2 Cl atoms in products)
- Hydrogen: \( x = 2 \) (since there are 2 H atoms in products)

### Step 4: Conclude the value of \( x \)
Since both the chlorine and hydrogen require \( x \) to be 2 for the equation to balance, we conclude:
\[ x = 2 \]

### Final Balanced Equation
The balanced equation is:
\[ \text{Na}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{CO}_2 + \text{H}_2\text{O} \]

### Answer
The value of \( x \) is 2.

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