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4. Balance the following equations using the algebraic method.

d. H₂SO₄ + NaOH → Na₂SO₄ + H₂о 4

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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: \[ \text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O} \]