To balance the given chemical equation:
[tex]\[ \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} \][/tex]
Let's follow these steps:
1. Write down the number of each type of atom on both sides of the equation:
- Reactants (Left Side):
- Na: 1 (from NaOH)
- O: 2 (1 from NaOH + 1 from H₂SO₄ which is included 4 Oxygens, but we should write it in such a way) (Rechecked)
- H: 3 (1 from NaOH + 2 from H₂SO₄)
- S: 1 (from H₂SO₄)
2. Balance the Sodium (Na) Atoms:
- We have 1 Sodium atom on the left side from NaOH, but on the right side, we have 2 Sodium atoms in Na₂SO₄.
- To equalize the number of Sodium atoms, we place a coefficient of 2 before NaOH on the left side:
[tex]\[ 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} \][/tex]
3. Balance the Hydrogen (H) Atoms:
- Now we have 2 Sodium atoms from 2 NaOH, so we also need to balance Hydrogens.
- There are 4 Hydrogen atoms on the left side after multiplying NaOH by 2 and having H₂SO₄.
- We need 2 Hydrogens for each water molecule, so we place a coefficient of 2 before H₂O on the right side:
[tex]\[ 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \][/tex]
4. Balance the Sulfur (S) and Oxygen (O) Atoms:
- We have now balanced the number of each atom.
- The equation showed 2 Oxygens (each molecule of NaOH), which compensates the 4 there from 2 molecules of water in products.
- 4 Sodium (2 NaOH each side) = 4 Sodiums (each atom in Na₂SO₄).
Thus, the balanced equation looks like:
[tex]\[ 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \][/tex]
So, the coefficient that goes in the blank for the balanced equation is:
[tex]\[ 2 \text{NaOH} + \underline{1}\text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \][/tex]
Thus, the coefficient for [tex]\( \text{H}_2\text{SO}_4 \)[/tex] in the balanced equation is 1.
