Answer :
Sure, let's balance the given chemical equations step-by-step.
### 2. Balancing the equation:
[tex]\[ \text{SO}_2 + \ldots \text{O}_2 \rightarrow \ldots \text{SO}_3 \][/tex]
1. Identify the reactants and products:
- Reactants: [tex]\(\text{SO}_2\)[/tex] and [tex]\(\text{O}_2\)[/tex]
- Products: [tex]\(\text{SO}_3\)[/tex]
2. Write down the unbalanced equation:
[tex]\[ \text{SO}_2 + \text{O}_2 \rightarrow \text{SO}_3 \][/tex]
3. Count the number of atoms for each element on both sides:
- Left side:
- S (Sulfur): 1 (from [tex]\(\text{SO}_2\)[/tex])
- O (Oxygen): 2 (from [tex]\(\text{SO}_2\)[/tex]) + 2 (from [tex]\(\text{O}_2\)[/tex]) = 4
- Right side:
- S (Sulfur): 1 (from [tex]\(\text{SO}_3\)[/tex])
- O (Oxygen): 3 (from [tex]\(\text{SO}_3\)[/tex])
4. Balance the sulfur (S) atoms:
- Sulfur atoms are already balanced: 1 on both sides.
5. Balance the oxygen (O) atoms:
- We have 4 oxygen atoms on the left side and 3 on the right side.
- Multiply [tex]\(\text{SO}_3\)[/tex] by 2 to get 6 oxygen atoms on the right:
[tex]\[ \text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3 \][/tex]
6. Balance remaining oxygen (O) atoms:
- Now, we have 4 oxygen atoms on the left (2 from [tex]\(\text{O}_2\)[/tex]) and 6 (from [tex]\(2\text{SO}_3\)[/tex]) on the right.
- To balance, multiply [tex]\(\text{SO}_2\)[/tex] by 2 and [tex]\(\text{O}_2\)[/tex] by 1.5:
[tex]\[ 2\text{SO}_2 + 1.5\text{O}_2 \rightarrow 2\text{SO}_3 \][/tex]
- However, coefficients should be whole numbers. Multiply everything by 2 to clear the fraction:
[tex]\[ 4\text{SO}_2 + 3\text{O}_2 \rightarrow 4\text{SO}_3 \][/tex]
Now, we see the simplest balanced equation is:
[tex]\[ 2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3 \][/tex]
Coefficients:
- [tex]\(\text{SO}_2: 2\)[/tex]
- [tex]\(\text{O}_2: 1\)[/tex]
- [tex]\(\text{SO}_3: 2\)[/tex]
### 3. Balancing the equation:
[tex]\[ \text{Na} + \_ \text{H}_2\text{O} \rightarrow \_ \text{NaOH} + \_ \text{H}_2 \][/tex]
1. Identify the reactants and products:
- Reactants: [tex]\(\text{Na}\)[/tex] and [tex]\(\text{H}_2\text{O}\)[/tex]
- Products: [tex]\(\text{NaOH}\)[/tex] and [tex]\(\text{H}_2\)[/tex]
2. Write down the unbalanced equation:
[tex]\[ \text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2 \][/tex]
3. Count the number of atoms for each element on both sides:
- Left side:
- Na (Sodium): 1 (from [tex]\(\text{Na}\)[/tex])
- H (Hydrogen): 2 (from [tex]\(\text{H}_2\text{O}\)[/tex])
- O (Oxygen): 1 (from [tex]\(\text{H}_2\text{O}\)[/tex])
- Right side:
- Na (Sodium): 1 (from [tex]\(\text{NaOH}\)[/tex])
- H (Hydrogen): 1 (from [tex]\(\text{NaOH}\)[/tex]) + 2 (from [tex]\(\text{H}_2\)[/tex]) = 3
- O (Oxygen): 1 (from [tex]\(\text{NaOH}\)[/tex])
4. Balance the sodium (Na) atoms:
- Sodium atoms are already balanced: 1 on both sides.
5. Balance the oxygen (O) atoms:
- Oxygen atoms are already balanced: 1 on both sides.
6. Balance the hydrogen (H) atoms:
- We have 2 hydrogen atoms on the left side and 3 hydrogen atoms on the right side.
- Multiply [tex]\(\text{H}_2\text{O}\)[/tex] by 2 and [tex]\(\text{NaOH}\)[/tex] by 2 to balance hydrogens and oxygens:
[tex]\[ 2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2 \][/tex]
Now we check:
- Left side:
- Na: 2 (from [tex]\(2\text{Na}\)[/tex])
- H: 4 (from [tex]\(2\text{H}_2\text{O}\)[/tex])
- O: 2 (from [tex]\(2\text{H}_2\text{O}\)[/tex])
- Right side:
- Na: 2 (from [tex]\(2\text{NaOH}\)[/tex])
- H: 4 (from [tex]\(2\text{NaOH}\)[/tex]) + 2 (from [tex]\(\text{H}_2\)[/tex]) = 4
- O: 2 (from [tex]\(2\text{NaOH}\)[/tex])
Final balanced equation:
[tex]\[ 2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2 \][/tex]
Coefficients:
- [tex]\(\text{Na}: 2\)[/tex]
- [tex]\(\text{H}_2\text{O}: 2\)[/tex]
- [tex]\(\text{NaOH}: 2\)[/tex]
- [tex]\(\text{H}_2: 1\)[/tex]
Therefore, the balanced coefficients for the equations are:
1. [tex]\(2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3\)[/tex]: [tex]\(\text{SO}_2: 2\)[/tex], [tex]\(\text{O}_2: 1\)[/tex], [tex]\(\text{SO}_3: 2\)[/tex].
2. [tex]\(2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2\)[/tex]: [tex]\(\text{Na}: 2\)[/tex], [tex]\(\text{H}_2\text{O}: 2\)[/tex], [tex]\(\text{NaOH}: 2\)[/tex], [tex]\(\text{H}_2: 1\)[/tex].
### 2. Balancing the equation:
[tex]\[ \text{SO}_2 + \ldots \text{O}_2 \rightarrow \ldots \text{SO}_3 \][/tex]
1. Identify the reactants and products:
- Reactants: [tex]\(\text{SO}_2\)[/tex] and [tex]\(\text{O}_2\)[/tex]
- Products: [tex]\(\text{SO}_3\)[/tex]
2. Write down the unbalanced equation:
[tex]\[ \text{SO}_2 + \text{O}_2 \rightarrow \text{SO}_3 \][/tex]
3. Count the number of atoms for each element on both sides:
- Left side:
- S (Sulfur): 1 (from [tex]\(\text{SO}_2\)[/tex])
- O (Oxygen): 2 (from [tex]\(\text{SO}_2\)[/tex]) + 2 (from [tex]\(\text{O}_2\)[/tex]) = 4
- Right side:
- S (Sulfur): 1 (from [tex]\(\text{SO}_3\)[/tex])
- O (Oxygen): 3 (from [tex]\(\text{SO}_3\)[/tex])
4. Balance the sulfur (S) atoms:
- Sulfur atoms are already balanced: 1 on both sides.
5. Balance the oxygen (O) atoms:
- We have 4 oxygen atoms on the left side and 3 on the right side.
- Multiply [tex]\(\text{SO}_3\)[/tex] by 2 to get 6 oxygen atoms on the right:
[tex]\[ \text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3 \][/tex]
6. Balance remaining oxygen (O) atoms:
- Now, we have 4 oxygen atoms on the left (2 from [tex]\(\text{O}_2\)[/tex]) and 6 (from [tex]\(2\text{SO}_3\)[/tex]) on the right.
- To balance, multiply [tex]\(\text{SO}_2\)[/tex] by 2 and [tex]\(\text{O}_2\)[/tex] by 1.5:
[tex]\[ 2\text{SO}_2 + 1.5\text{O}_2 \rightarrow 2\text{SO}_3 \][/tex]
- However, coefficients should be whole numbers. Multiply everything by 2 to clear the fraction:
[tex]\[ 4\text{SO}_2 + 3\text{O}_2 \rightarrow 4\text{SO}_3 \][/tex]
Now, we see the simplest balanced equation is:
[tex]\[ 2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3 \][/tex]
Coefficients:
- [tex]\(\text{SO}_2: 2\)[/tex]
- [tex]\(\text{O}_2: 1\)[/tex]
- [tex]\(\text{SO}_3: 2\)[/tex]
### 3. Balancing the equation:
[tex]\[ \text{Na} + \_ \text{H}_2\text{O} \rightarrow \_ \text{NaOH} + \_ \text{H}_2 \][/tex]
1. Identify the reactants and products:
- Reactants: [tex]\(\text{Na}\)[/tex] and [tex]\(\text{H}_2\text{O}\)[/tex]
- Products: [tex]\(\text{NaOH}\)[/tex] and [tex]\(\text{H}_2\)[/tex]
2. Write down the unbalanced equation:
[tex]\[ \text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2 \][/tex]
3. Count the number of atoms for each element on both sides:
- Left side:
- Na (Sodium): 1 (from [tex]\(\text{Na}\)[/tex])
- H (Hydrogen): 2 (from [tex]\(\text{H}_2\text{O}\)[/tex])
- O (Oxygen): 1 (from [tex]\(\text{H}_2\text{O}\)[/tex])
- Right side:
- Na (Sodium): 1 (from [tex]\(\text{NaOH}\)[/tex])
- H (Hydrogen): 1 (from [tex]\(\text{NaOH}\)[/tex]) + 2 (from [tex]\(\text{H}_2\)[/tex]) = 3
- O (Oxygen): 1 (from [tex]\(\text{NaOH}\)[/tex])
4. Balance the sodium (Na) atoms:
- Sodium atoms are already balanced: 1 on both sides.
5. Balance the oxygen (O) atoms:
- Oxygen atoms are already balanced: 1 on both sides.
6. Balance the hydrogen (H) atoms:
- We have 2 hydrogen atoms on the left side and 3 hydrogen atoms on the right side.
- Multiply [tex]\(\text{H}_2\text{O}\)[/tex] by 2 and [tex]\(\text{NaOH}\)[/tex] by 2 to balance hydrogens and oxygens:
[tex]\[ 2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2 \][/tex]
Now we check:
- Left side:
- Na: 2 (from [tex]\(2\text{Na}\)[/tex])
- H: 4 (from [tex]\(2\text{H}_2\text{O}\)[/tex])
- O: 2 (from [tex]\(2\text{H}_2\text{O}\)[/tex])
- Right side:
- Na: 2 (from [tex]\(2\text{NaOH}\)[/tex])
- H: 4 (from [tex]\(2\text{NaOH}\)[/tex]) + 2 (from [tex]\(\text{H}_2\)[/tex]) = 4
- O: 2 (from [tex]\(2\text{NaOH}\)[/tex])
Final balanced equation:
[tex]\[ 2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2 \][/tex]
Coefficients:
- [tex]\(\text{Na}: 2\)[/tex]
- [tex]\(\text{H}_2\text{O}: 2\)[/tex]
- [tex]\(\text{NaOH}: 2\)[/tex]
- [tex]\(\text{H}_2: 1\)[/tex]
Therefore, the balanced coefficients for the equations are:
1. [tex]\(2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3\)[/tex]: [tex]\(\text{SO}_2: 2\)[/tex], [tex]\(\text{O}_2: 1\)[/tex], [tex]\(\text{SO}_3: 2\)[/tex].
2. [tex]\(2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2\)[/tex]: [tex]\(\text{Na}: 2\)[/tex], [tex]\(\text{H}_2\text{O}: 2\)[/tex], [tex]\(\text{NaOH}: 2\)[/tex], [tex]\(\text{H}_2: 1\)[/tex].
