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Balancing chemical equations involves making sure the number of atoms for each element is equal on both the reactant and product sides of the equation. Let's balance the given chemical equation:
[tex]\[ \text{Unbalanced Equation: } \text{NH}_4\text{OH} + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4 + \text{H}_2\text{O} \][/tex]
### Step 1: List the number of atoms of each element on both sides.
Reactants:
- NH[tex]\(_4\)[/tex]OH: 1 N, 5 H (4 from NH[tex]\(_4\)[/tex] and 1 from OH), 1 O
- H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]: 2 H, 1 S, 4 O
Products:
- (NH[tex]\(_4\)[/tex])[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]: 2 N, 8 H (4 from each NH[tex]\(_4\)[/tex]), 1 S, 4 O
- H[tex]\(_2\)[/tex]O: 2 H, 1 O
### Step 2: Start balancing with one element at a time.
#### Nitrogen (N)
On the left side, we have 1 N in NH[tex]\(_4\)[/tex]OH. On the right side, we have 2 N in (NH[tex]\(_4\)[/tex])[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]. So we need 2 NH[tex]\(_4\)[/tex]OH to get 2 N atoms.
[tex]\[ 2 \text{NH}_4\text{OH} + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4 + \text{H}_2\text{O} \][/tex]
Reactants:
- 2 NH[tex]\(_4\)[/tex]OH: 2 N, 10 H (8 from NH[tex]\(_4\)[/tex] and 2 from OH), 2 O
- H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]: 2 H, 1 S, 4 O
Products:
- (NH[tex]\(_4\)[/tex])[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]: 2 N, 8 H (4 from each NH[tex]\(_4\)[/tex]), 1 S, 4 O
- H[tex]\(_2\)[/tex]O: 2 H, 1 O
### Step 3: Balance the remaining elements.
#### Hydrogen (H)
On the left side, we now have 2 NH[tex]\(_4\)[/tex]OH contributing 10 H atoms (8 from NH[tex]\(_4\)[/tex] and 2 from OH), and H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] contributing 2 H atoms, totaling 12 H atoms.
On the right side, (NH[tex]\(_4\)[/tex])[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] contributes 8 H atoms, and each H[tex]\(_2\)[/tex]O contributes 2 H atoms. To balance, we need 4 H atoms in water (total of 12 H atoms).
That means we need:
[tex]\[ 2 \text{H}_2\text{O} \][/tex]
on the right side.
### Step 4: Confirm all elements are balanced.
Final balanced equation:
[tex]\[ 2 \text{NH}_4\text{OH} + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4 + 2 \text{H}_2\text{O} \][/tex]
Reactants:
- 2 NH[tex]\(_4\)[/tex]OH: 2 N, 10 H, 2 O
- H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]: 2 H, 1 S, 4 O
Total: 2 N, 12 H, 1 S, 6 O
Products:
- (NH[tex]\(_4\)[/tex])[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]: 2 N, 8 H, 1 S, 4 O
- 2 H[tex]\(_2\)[/tex]O: 4 H, 2 O
Total: 2 N, 12 H, 1 S, 6 O
The final balanced chemical equation is:
[tex]\[ 2 \text{NH}_4\text{OH} + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4 + 2 \text{H}_2\text{O} \][/tex]
This ensures that the number of each type of atom is the same on both sides of the equation.
[tex]\[ \text{Unbalanced Equation: } \text{NH}_4\text{OH} + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4 + \text{H}_2\text{O} \][/tex]
### Step 1: List the number of atoms of each element on both sides.
Reactants:
- NH[tex]\(_4\)[/tex]OH: 1 N, 5 H (4 from NH[tex]\(_4\)[/tex] and 1 from OH), 1 O
- H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]: 2 H, 1 S, 4 O
Products:
- (NH[tex]\(_4\)[/tex])[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]: 2 N, 8 H (4 from each NH[tex]\(_4\)[/tex]), 1 S, 4 O
- H[tex]\(_2\)[/tex]O: 2 H, 1 O
### Step 2: Start balancing with one element at a time.
#### Nitrogen (N)
On the left side, we have 1 N in NH[tex]\(_4\)[/tex]OH. On the right side, we have 2 N in (NH[tex]\(_4\)[/tex])[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]. So we need 2 NH[tex]\(_4\)[/tex]OH to get 2 N atoms.
[tex]\[ 2 \text{NH}_4\text{OH} + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4 + \text{H}_2\text{O} \][/tex]
Reactants:
- 2 NH[tex]\(_4\)[/tex]OH: 2 N, 10 H (8 from NH[tex]\(_4\)[/tex] and 2 from OH), 2 O
- H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]: 2 H, 1 S, 4 O
Products:
- (NH[tex]\(_4\)[/tex])[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]: 2 N, 8 H (4 from each NH[tex]\(_4\)[/tex]), 1 S, 4 O
- H[tex]\(_2\)[/tex]O: 2 H, 1 O
### Step 3: Balance the remaining elements.
#### Hydrogen (H)
On the left side, we now have 2 NH[tex]\(_4\)[/tex]OH contributing 10 H atoms (8 from NH[tex]\(_4\)[/tex] and 2 from OH), and H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] contributing 2 H atoms, totaling 12 H atoms.
On the right side, (NH[tex]\(_4\)[/tex])[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] contributes 8 H atoms, and each H[tex]\(_2\)[/tex]O contributes 2 H atoms. To balance, we need 4 H atoms in water (total of 12 H atoms).
That means we need:
[tex]\[ 2 \text{H}_2\text{O} \][/tex]
on the right side.
### Step 4: Confirm all elements are balanced.
Final balanced equation:
[tex]\[ 2 \text{NH}_4\text{OH} + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4 + 2 \text{H}_2\text{O} \][/tex]
Reactants:
- 2 NH[tex]\(_4\)[/tex]OH: 2 N, 10 H, 2 O
- H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]: 2 H, 1 S, 4 O
Total: 2 N, 12 H, 1 S, 6 O
Products:
- (NH[tex]\(_4\)[/tex])[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex]: 2 N, 8 H, 1 S, 4 O
- 2 H[tex]\(_2\)[/tex]O: 4 H, 2 O
Total: 2 N, 12 H, 1 S, 6 O
The final balanced chemical equation is:
[tex]\[ 2 \text{NH}_4\text{OH} + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4 + 2 \text{H}_2\text{O} \][/tex]
This ensures that the number of each type of atom is the same on both sides of the equation.
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