Answer :
Sure, let's go through the steps to balance the given chemical equations one by one in detail.
(i) Balancing the equation: [tex]\( \text{CO}_2 + \text{H}_2\text{O} \longrightarrow \text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2 \)[/tex]
1. Balance carbon (C) atoms:
- On the right side, there are 6 carbon atoms in [tex]\( \text{C}_6\text{H}_{12}\text{O}_6 \)[/tex].
- Therefore, we need 6 [tex]\( \text{CO}_2 \)[/tex] molecules on the left side.
[tex]\[ 6 \text{CO}_2 + \text{H}_2\text{O} \longrightarrow \text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2 \][/tex]
2. Balance hydrogen (H) atoms:
- On the right side, there are 12 hydrogen atoms in [tex]\( \text{C}_6\text{H}_{12}\text{O}_6 \)[/tex].
- Therefore, we need 6 [tex]\( \text{H}_2\text{O} \)[/tex] molecules on the left side.
[tex]\[ 6 \text{CO}_2 + 6 \text{H}_2\text{O} \longrightarrow \text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2 \][/tex]
3. Balance oxygen (O) atoms:
- On the left side, we have:
- [tex]\(6 \text{CO}_2 \)[/tex] which gives [tex]\(6 \times 2 = 12\)[/tex] oxygen atoms.
- [tex]\(6 \text{H}_2\text{O} \)[/tex] which gives [tex]\(6 \times 1 = 6\)[/tex] oxygen atoms.
- In total, there are 18 oxygen atoms on the left side.
- On the right side, we have:
- [tex]\( \text{C}_6\text{H}_{12}\text{O}_6 \)[/tex] which contains 6 oxygen atoms.
- To balance the remaining 12 oxygen atoms, we need 6 [tex]\( \text{O}_2 \)[/tex] molecules, each having 2 oxygen atoms [tex]\( (6 \times 2 = 12) \)[/tex].
Thus, the balanced equation is:
[tex]\[ 6 \text{CO}_2 + 6 \text{H}_2\text{O} \longrightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6 \text{O}_2 \][/tex]
(ii) Balancing the equation: [tex]\( \text{Fe} + \text{O}_2 \longrightarrow \text{Fe}_2\text{O}_3 \)[/tex]
1. Balance iron (Fe) atoms:
- On the right side, there are 2 iron atoms in [tex]\( \text{Fe}_2\text{O}_3 \)[/tex].
- Therefore, we need 2 [tex]\( \text{Fe} \)[/tex] atoms on the left side.
[tex]\[ 2 \text{Fe} + \text{O}_2 \longrightarrow \text{Fe}_2\text{O}_3 \][/tex]
2. Balance oxygen (O) atoms:
- On the right side, there are 3 oxygen atoms in [tex]\( \text{Fe}_2\text{O}_3 \)[/tex].
- To balance this, we need 1.5 [tex]\( \text{O}_2 \)[/tex] molecules on the left side, as each [tex]\( \text{O}_2 \)[/tex] molecule has 2 oxygen atoms [tex]\( (1.5 \times 2 = 3) \)[/tex].
[tex]\[ 2 \text{Fe} + 1.5 \text{O}_2 \longrightarrow \text{Fe}_2\text{O}_3 \][/tex]
3. Eliminate fractions:
- To eliminate the fraction, we multiply all coefficients in the equation by 2:
[tex]\[ 2 \times 2 \text{Fe} + 2 \times 1.5 \text{O}_2 \longrightarrow 2 \times \text{Fe}_2\text{O}_3 \][/tex]
- This results in:
[tex]\[ 4 \text{Fe} + 3 \text{O}_2 \longrightarrow 2 \text{Fe}_2\text{O}_3 \][/tex]
Thus, the balanced equation is:
[tex]\[ 4 \text{Fe} + 3 \text{O}_2 \longrightarrow 2 \text{Fe}_2\text{O}_3 \][/tex]
To summarize, the balanced chemical equations are:
1. [tex]\( 6 \text{CO}_2 + 6 \text{H}_2\text{O} \longrightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6 \text{O}_2 \)[/tex]
2. [tex]\( 4 \text{Fe} + 3 \text{O}_2 \longrightarrow 2 \text{Fe}_2\text{O}_3 \)[/tex]
(i) Balancing the equation: [tex]\( \text{CO}_2 + \text{H}_2\text{O} \longrightarrow \text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2 \)[/tex]
1. Balance carbon (C) atoms:
- On the right side, there are 6 carbon atoms in [tex]\( \text{C}_6\text{H}_{12}\text{O}_6 \)[/tex].
- Therefore, we need 6 [tex]\( \text{CO}_2 \)[/tex] molecules on the left side.
[tex]\[ 6 \text{CO}_2 + \text{H}_2\text{O} \longrightarrow \text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2 \][/tex]
2. Balance hydrogen (H) atoms:
- On the right side, there are 12 hydrogen atoms in [tex]\( \text{C}_6\text{H}_{12}\text{O}_6 \)[/tex].
- Therefore, we need 6 [tex]\( \text{H}_2\text{O} \)[/tex] molecules on the left side.
[tex]\[ 6 \text{CO}_2 + 6 \text{H}_2\text{O} \longrightarrow \text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2 \][/tex]
3. Balance oxygen (O) atoms:
- On the left side, we have:
- [tex]\(6 \text{CO}_2 \)[/tex] which gives [tex]\(6 \times 2 = 12\)[/tex] oxygen atoms.
- [tex]\(6 \text{H}_2\text{O} \)[/tex] which gives [tex]\(6 \times 1 = 6\)[/tex] oxygen atoms.
- In total, there are 18 oxygen atoms on the left side.
- On the right side, we have:
- [tex]\( \text{C}_6\text{H}_{12}\text{O}_6 \)[/tex] which contains 6 oxygen atoms.
- To balance the remaining 12 oxygen atoms, we need 6 [tex]\( \text{O}_2 \)[/tex] molecules, each having 2 oxygen atoms [tex]\( (6 \times 2 = 12) \)[/tex].
Thus, the balanced equation is:
[tex]\[ 6 \text{CO}_2 + 6 \text{H}_2\text{O} \longrightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6 \text{O}_2 \][/tex]
(ii) Balancing the equation: [tex]\( \text{Fe} + \text{O}_2 \longrightarrow \text{Fe}_2\text{O}_3 \)[/tex]
1. Balance iron (Fe) atoms:
- On the right side, there are 2 iron atoms in [tex]\( \text{Fe}_2\text{O}_3 \)[/tex].
- Therefore, we need 2 [tex]\( \text{Fe} \)[/tex] atoms on the left side.
[tex]\[ 2 \text{Fe} + \text{O}_2 \longrightarrow \text{Fe}_2\text{O}_3 \][/tex]
2. Balance oxygen (O) atoms:
- On the right side, there are 3 oxygen atoms in [tex]\( \text{Fe}_2\text{O}_3 \)[/tex].
- To balance this, we need 1.5 [tex]\( \text{O}_2 \)[/tex] molecules on the left side, as each [tex]\( \text{O}_2 \)[/tex] molecule has 2 oxygen atoms [tex]\( (1.5 \times 2 = 3) \)[/tex].
[tex]\[ 2 \text{Fe} + 1.5 \text{O}_2 \longrightarrow \text{Fe}_2\text{O}_3 \][/tex]
3. Eliminate fractions:
- To eliminate the fraction, we multiply all coefficients in the equation by 2:
[tex]\[ 2 \times 2 \text{Fe} + 2 \times 1.5 \text{O}_2 \longrightarrow 2 \times \text{Fe}_2\text{O}_3 \][/tex]
- This results in:
[tex]\[ 4 \text{Fe} + 3 \text{O}_2 \longrightarrow 2 \text{Fe}_2\text{O}_3 \][/tex]
Thus, the balanced equation is:
[tex]\[ 4 \text{Fe} + 3 \text{O}_2 \longrightarrow 2 \text{Fe}_2\text{O}_3 \][/tex]
To summarize, the balanced chemical equations are:
1. [tex]\( 6 \text{CO}_2 + 6 \text{H}_2\text{O} \longrightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6 \text{O}_2 \)[/tex]
2. [tex]\( 4 \text{Fe} + 3 \text{O}_2 \longrightarrow 2 \text{Fe}_2\text{O}_3 \)[/tex]
