Question

4. Balance the following equations using the algebraic method.

d. H₂SO₄ + NaOH → Na₂SO₄ + H₂о 4

Answer 1

To balance the equation \( \text{H}_2\text{SO}_4 + \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} \) using the algebraic method, we will assign a variable to each compound and then create equations based on the number of atoms of each element.

Let's assign:
- \( a \) to \( \text{H}_2\text{SO}_4 \)
- \( b \) to \( \text{NaOH} \)
- \( c \) to \( \text{Na}_2\text{SO}_4 \)
- \( d \) to \( \text{H}_2\text{O} \)

This gives us the following unbalanced equation:
\[ a \text{H}_2\text{SO}_4 + b \text{NaOH} \rightarrow c \text{Na}_2\text{SO}_4 + d \text{H}_2\text{O} \]

Now we count the atoms for each element on both sides of the equation:

1. **Sulfur (S)**:
- Left: \( a \) (from \( \text{H}_2\text{SO}_4 \))
- Right: \( c \) (from \( \text{Na}_2\text{SO}_4 \))

Equation for sulfur:
\( a = c \)

2. **Sodium (Na)**:
- Left: \( b \) (from \( \text{NaOH} \))
- Right: \( 2c \) (from \( \text{Na}_2\text{SO}_4 \))

Equation for sodium:
\( b = 2c \)

3. **Hydrogen (H)**:
- Left: \( 2a + b \) (from \( \text{H}_2\text{SO}_4 \) and \( \text{NaOH} \))
- Right: \( 2d \) (from \( \text{H}_2\text{O} \))

Equation for hydrogen:
\( 2a + b = 2d \)

4. **Oxygen (O)**:
- Left: \( 4a + b \) (from \( \text{H}_2\text{SO}_4 \) and \( \text{NaOH} \))
- Right: \( 4c + d \) (from \( \text{Na}_2\text{SO}_4 \) and \( \text{H}_2\text{O} \))

Equation for oxygen:
\( 4a + b = 4c + d \)

Now we have a system of equations:
1. \( a = c \)
2. \( b = 2c \)
3. \( 2a + b = 2d \)
4. \( 4a + b = 4c + d \)

Now let's solve this system step by step.

Starting with \( c = a \) from equation (1) and \( b = 2c = 2a \) from equation (2), we can substitute into equations (3) and (4):

**Substituting \( b = 2a \) into equation (3):**

\[
2a + 2a = 2d \quad \Rightarrow \quad 4a = 2d \quad \Rightarrow \quad d = 2a
\]

**Substituting \( b = 2a \) and \( c = a \) into equation (4):**

\[
4a + 2a = 4a + d
\]
Substituting \( d = 2a \):

\[
4a + 2a = 4a + 2a \quad \Rightarrow \quad 6a = 6a
\]
This is always true, confirming our findings.

Now we can assign a value to \( a \). Let’s set \( a = 1 \):

- Then, \( c = a = 1 \)
- \( b = 2a = 2 \)
- \( d = 2a = 2 \)

Now we can rewrite the coefficients:

The balanced equation is:
\[
1 \text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow 1 \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O}
\]
or simply:
\[
\text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O}
\]

Thus, the final balanced equation is:
\[
\text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O}
\]