Sodium hydroxide (NaOH) reacts with sulfuric acid (H2SO4) to form sodium sulfate (Na2SO4) and water (H2O).

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Sodium hydroxide (NaOH) reacts with sulfuric acid (H2SO4) to form sodium sulfate (Na2SO4) and water (H2O). The unbalanced chemical equation is NaOH + H2SO4 → Na2SO4 + H2O. To balance this equation, unknowns (x, y, z, and w) are inserted: xNaOH + yH2SO4 → zNa2SO4 + wH2O. Next, compare the number of sodium (Na), oxygen (O), hydrogen (H), and sulfur (S) atoms on the left side with the number on the right side. This gives four linear equations. Rewrite these equations in standard form: a1x + a2y + a3z + a4w = 0, a5x + a6y + a7z + a8w = 0, a9x + a10y + a11z + a12w = 0, a13x + a14y + a15z + a16w = 0. Write this system in matrix form (augmented matrix) and use row operations to reduce it to row echelon form. Find the values of the unknowns (x, y, and w). Since we are dealing with atoms, it is convenient to choose values so that all the unknowns are positive integers.

The Correct Answer and Explanation is:

Let’s go step by step to balance the chemical equation for the reaction of sodium hydroxide (NaOH) with sulfuric acid (H2SO4) to form sodium sulfate (Na2SO4) and water (H2O).

Unbalanced Equation:

NaOH+H2SO4→Na2SO4+H2O\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}NaOH+H2​SO4​→Na2​SO4​+H2​O

Step 1: Introduce unknowns

Let the coefficients of the compounds be represented by unknowns xxx, yyy, zzz, and www for NaOH, H2SO4, Na2SO4, and H2O, respectively:xNaOH+yH2SO4→zNa2SO4+wH2Ox\text{NaOH} + y\text{H}_2\text{SO}_4 \rightarrow z\text{Na}_2\text{SO}_4 + w\text{H}_2\text{O}xNaOH+yH2​SO4​→zNa2​SO4​+wH2​O

Step 2: Balance the elements

  • Sodium (Na): On the left side, sodium is only in NaOH. On the right side, sodium is in Na2SO4. Since Na2SO4 contains 2 Na atoms, we need to have x=2zx = 2zx=2z.
  • Sulfur (S): Sulfur is only in H2SO4 on the left side and Na2SO4 on the right side. Since Na2SO4 contains 1 sulfur atom, we need y=zy = zy=z.
  • Oxygen (O): Oxygen appears in NaOH, H2SO4, Na2SO4, and H2O. On the left side, oxygen is present in NaOH (1 per molecule) and H2SO4 (4 per molecule). On the right side, oxygen is present in Na2SO4 (4 per molecule) and H2O (1 per molecule). So, x+4y=4z+wx + 4y = 4z + wx+4y=4z+w.
  • Hydrogen (H): Hydrogen is present in NaOH (1 per molecule) and H2SO4 (2 per molecule) on the left side. On the right side, hydrogen is in H2O (2 per molecule). So, x+2y=2wx + 2y = 2wx+2y=2w.

Step 3: Set up the system of equations

From the above, we have the following system of equations:

  1. x=2zx = 2zx=2z (Sodium balance)
  2. y=zy = zy=z (Sulfur balance)
  3. x+4y=4z+wx + 4y = 4z + wx+4y=4z+w (Oxygen balance)
  4. x+2y=2wx + 2y = 2wx+2y=2w (Hydrogen balance)

Now, substitute x=2zx = 2zx=2z and y=zy = zy=z into the other equations.

Substitute into the third equation:2z+4z=4z+w⇒6z=4z+w⇒w=2z2z + 4z = 4z + w \quad \Rightarrow \quad 6z = 4z + w \quad \Rightarrow \quad w = 2z2z+4z=4z+w⇒6z=4z+w⇒w=2z

Substitute into the fourth equation:2z+2z=2w⇒4z=2w⇒w=2z2z + 2z = 2w \quad \Rightarrow \quad 4z = 2w \quad \Rightarrow \quad w = 2z2z+2z=2w⇒4z=2w⇒w=2z

Step 4: Final values for xxx, yyy, and www

From the equations, we have:

  • x=2zx = 2zx=2z
  • y=zy = zy=z
  • w=2zw = 2zw=2z

Since zzz must be a positive integer, let’s choose z=1z = 1z=1 (the smallest positive integer). This gives:

  • x=2x = 2x=2
  • y=1y = 1y=1
  • w=2w = 2w=2

Step 5: Write the balanced equation

Now, substitute these values back into the chemical equation:2NaOH+H2SO4→Na2SO4+2H2O2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}2NaOH+H2​SO4​→Na2​SO4​+2H2​O

Final Answer:

The balanced chemical equation is:2NaOH+H2SO4→Na2SO4+2H2O2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}2NaOH+H2​SO4​→Na2​SO4​+2H2​O

Step 6: Matrix Form and Row Operations

Now we set up the matrix system:(10−2001−1014−41120−2)(xyzw)=(0000)\begin{pmatrix} 1 & 0 & -2 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 4 & -4 & 1 \\ 1 & 2 & 0 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}​1011​0142​−2−1−40​001−2​​​xyzw​​=​0000​​

This matrix system can be solved using row reduction techniques, yielding the values x=2x = 2x=2, y=1y = 1y=1, and w=2w = 2w=2, as derived.

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