In the following equations
`Na_2CO_3 + xHCl to 2NaCl + CO_2 + H_2O`
the value of x is

Options

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

Explanation

To find the value of \( x \) in the equation

\[ \text{Na}_2\text{CO}_3 + x\text{HCl} \rightarrow 2\text{NaCl} + \text{CO}_2 + \text{H}_2\text{O} \]

we need to balance the chemical equation. Here are the steps to do that:

### 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: Balance the equation
From the above counts, we can see:
- Sodium is already balanced (2 on both sides).
- Carbon is balanced (1 on both sides).
- Oxygen is balanced (3 on both sides).
- For Hydrogen and Chlorine, we need to balance them.

To balance the hydrogen and chlorine, we note that there are 2 hydrogen atoms in the products (from \(\text{H}_2\text{O}\)) and 2 chlorine atoms (from \(2\text{NaCl}\)). Therefore, we need to set \(x\) to 2:

\[ x = 2 \]

### Step 4: Write the balanced equation
The balanced equation is:
\[ \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{CO}_2 + \text{H}_2\text{O} \]

### Conclusion
Thus, the value of \( x \) is 2.

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