Question

Balance
__Na2CO3 + __HCl->__NaCl+CO2+__H2O

Answer 1

To balance the chemical equation \( \text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{CO}_2 + \text{H}_2\text{O} \), follow these steps:

1. **List the number of each type of atom on both sides of the equation:**
- Reactants:
- Na: 2 (from Na₂CO₃)
- C: 1 (from Na₂CO₃)
- O: 3 (from Na₂CO₃)
- H: 1 (from HCl)
- Cl: 1 (from HCl)

- Products:
- Na: 1 (from NaCl)
- C: 1 (from CO₂)
- O: 2 (from CO₂) + 1 (from H₂O) = 3
- H: 2 (from H₂O)
- Cl: 1 (from NaCl)

2. **Balance Sodium (\( \text{Na} \))**:
- There are 2 sodium atoms in the reactants, so we need 2 NaCl in the products:
\[
\text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow 2 \text{NaCl} + \text{CO}_2 + \text{H}_2\text{O}
\]

3. **Rewrite and update the counts**:
- Now we have:
- Reactants:
- \( \text{Na}_2\text{CO}_3 \): 2 Na, 1 C, 3 O, 1 H, 1 Cl
- \( \text{HCl} \): 1 H, 1 Cl
- Products:
- \( 2 \text{NaCl} \): 2 Na, 2 Cl
- \( \text{CO}_2 \): 1 C, 2 O
- \( \text{H}_2\text{O} \): 2 H, 1 O

4. **Balance Chlorine (\( \text{Cl} \))**:
- Since we now have 2 Cl in the products from 2 NaCl, we need 2 HCl in the reactants:
\[
\text{Na}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{CO}_2 + \text{H}_2\text{O}
\]

5. **Update the counts once more:**
- Now, ensuring everything is balanced:
- Reactants:
- \( \text{Na}_2\text{CO}_3 \): 2 Na, 1 C, 3 O
- \( 2 \text{HCl} \): 2 H, 2 Cl
- Products:
- \( 2 \text{NaCl} \): 2 Na, 2 Cl
- \( \text{CO}_2 \): 1 C, 2 O
- \( \text{H}_2\text{O} \): 2 H, 1 O

6. **Count the oxygen (O)**:
- In the reactants: 3 O
- In the products: \( 2 \text{O} \text{ (from } CO_2\text{)} + 1 \text{O (from } H_2O\text{)} = 3 \text{ O}\)

Now everything is balanced.

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