Answer :
To balance the chemical equation [tex]\( \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} \)[/tex], we need to ensure that the number of atoms of each element on the reactant side matches the number of atoms of each element on the product side. Let's go through the steps to balance this equation step-by-step.
1. Identify the number of atoms of each element on both sides:
- Reactants:
- Na: 1 (from NaOH)
- O: 5 (from NaOH and H_2SO_4, i.e., 1 from NaOH and 4 from H_2SO_4)
- H: 3 (from NaOH and H_2SO_4, i.e., 1 from NaOH and 2 from H_2SO_4)
- S: 1 (from H_2SO_4)
- Products:
- Na: 2 (from Na_2SO_4)
- O: 5 (from Na_2SO_4 and H_2O, i.e., 4 from Na_2SO_4 and 1 from H_2O)
- H: 2 (from H_2O)
- S: 1 (from Na_2SO_4)
2. Balance the sodium (Na) atoms:
- There is 1 Na atom on the reactant side and 2 Na atoms on the product side. Balance Na by placing a coefficient of 2 in front of NaOH:
[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:
- On the reactant side, we now have 2 Na atoms (from 2 molecules of NaOH) plus 2 more H atoms (from H_2SO_4), totaling 4 H atoms. On the product side, there are 2 H atoms (from H_2O). Thus, place a coefficient of 2 in front of H_2O:
[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 oxygen (O) atoms:
- On the reactant side, we have 2 O atoms (from 2 molecules of NaOH) plus 4 more O atoms (from H_2SO_4), totaling 6 O atoms.
- On the product side, we now have 4 O atoms (from Na_2SO_4) plus 2 more O atoms (from 2 molecules of H_2O), totaling 6 O atoms.
- Therefore, the oxygen atoms are balanced.
5. Verify that all elements are balanced:
- Reactant side: 2 Na, 1 S, 6 O, 4 H
- Product side: 2 Na, 1 S, 6 O, 4 H
All elements are balanced, and the final balanced equation is:
[tex]\[ 2 \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O} \][/tex]
The coefficient that goes in the blank is:
[tex]\[ 2 \][/tex]
1. Identify the number of atoms of each element on both sides:
- Reactants:
- Na: 1 (from NaOH)
- O: 5 (from NaOH and H_2SO_4, i.e., 1 from NaOH and 4 from H_2SO_4)
- H: 3 (from NaOH and H_2SO_4, i.e., 1 from NaOH and 2 from H_2SO_4)
- S: 1 (from H_2SO_4)
- Products:
- Na: 2 (from Na_2SO_4)
- O: 5 (from Na_2SO_4 and H_2O, i.e., 4 from Na_2SO_4 and 1 from H_2O)
- H: 2 (from H_2O)
- S: 1 (from Na_2SO_4)
2. Balance the sodium (Na) atoms:
- There is 1 Na atom on the reactant side and 2 Na atoms on the product side. Balance Na by placing a coefficient of 2 in front of NaOH:
[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:
- On the reactant side, we now have 2 Na atoms (from 2 molecules of NaOH) plus 2 more H atoms (from H_2SO_4), totaling 4 H atoms. On the product side, there are 2 H atoms (from H_2O). Thus, place a coefficient of 2 in front of H_2O:
[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 oxygen (O) atoms:
- On the reactant side, we have 2 O atoms (from 2 molecules of NaOH) plus 4 more O atoms (from H_2SO_4), totaling 6 O atoms.
- On the product side, we now have 4 O atoms (from Na_2SO_4) plus 2 more O atoms (from 2 molecules of H_2O), totaling 6 O atoms.
- Therefore, the oxygen atoms are balanced.
5. Verify that all elements are balanced:
- Reactant side: 2 Na, 1 S, 6 O, 4 H
- Product side: 2 Na, 1 S, 6 O, 4 H
All elements are balanced, and the final balanced equation is:
[tex]\[ 2 \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O} \][/tex]
The coefficient that goes in the blank is:
[tex]\[ 2 \][/tex]
