To write the chemical equations for the combustion reactions of the given hydrocarbons, we need to follow a systematic approach. Combustion reactions typically involve a hydrocarbon reacting with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O). The general form of a combustion reaction can be represented as:
\[ \text{Hydrocarbon} + O_2 \rightarrow CO_2 + H_2O \]
Now, let's write the balanced chemical equations for each of the hydrocarbons given in the question.
### (i) Combustion of Butane (C₄H₁₀)
1. **Write the unbalanced equation:**
\[ C_4H_{10} + O_2 \rightarrow CO_2 + H_2O \]
2. **Balance the carbon atoms:**
There are 4 carbon atoms in butane, so we need 4 CO₂.
\[ C_4H_{10} + O_2 \rightarrow 4CO_2 + H_2O \]
3. **Balance the hydrogen atoms:**
There are 10 hydrogen atoms in butane, so we need 5 H₂O.
\[ C_4H_{10} + O_2 \rightarrow 4CO_2 + 5H_2O \]
4. **Balance the oxygen atoms:**
On the right side, we have \(4 \times 2 = 8\) from CO₂ and \(5 \times 1 = 5\) from H₂O, totaling 13 oxygen atoms. Therefore, we need \( \frac{13}{2} \) O₂ on the left side.
\[ C_4H_{10} + \frac{13}{2} O_2 \rightarrow 4CO_2 + 5H_2O \]
5. **Multiply through by 2 to eliminate the fraction:**
\[ 2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O \]
### (ii) Combustion of Pentene (C₅H₁₀)
1. **Write the unbalanced equation:**
\[ C_5H_{10} + O_2 \rightarrow CO_2 + H_2O \]
2. **Balance the carbon atoms:**
There are 5 carbon atoms in pentene, so we need 5 CO₂.
\[ C_5H_{10} + O_2 \rightarrow 5CO_2 + H_2O \]
3. **Balance the hydrogen atoms:**
There are 10 hydrogen atoms in pentene, so we need 5 H₂O.
\[ C_5H_{10} + O_2 \rightarrow 5CO_2 + 5H_2O \]
4. **Balance the oxygen atoms:**
On the right side, we have \(5 \times 2 = 10\) from CO₂ and \(5 \times 1 = 5\) from H₂O, totaling 15 oxygen atoms. Therefore, we need \( \frac{15}{2} \) O₂ on the left side.
\[ C_5H_{10} + \frac{15}{2} O_2 \rightarrow 5CO_2 + 5H_2O \]
5. **Multiply through by 2 to eliminate the fraction:**
\[ 2C_5H_{10} + 15O_2 \rightarrow 10CO_2 + 10H_2O \]
### (iii) Combustion of Hexyne (C₆H₁₀)
1. **Write the unbalanced equation:**
\[ C_6H_{10} + O_2 \rightarrow CO_2 + H_2O \]
2. **Balance the carbon atoms:**
There are 6 carbon atoms in hexyne, so we need 6 CO₂.
\[ C_6H_{10} + O_2 \rightarrow 6CO_2 + H_2O \]
3. **Balance the hydrogen atoms:**
There are 10 hydrogen atoms in hexyne, so we need 5 H₂O.
\[ C_6H_{10} + O_2 \rightarrow 6CO_2 + 5H_2O \]
4. **Balance the oxygen atoms:**
On the right side, we have \(6 \times 2 = 12\) from CO₂ and \(5 \times 1 = 5\) from H₂O, totaling 17 oxygen atoms. Therefore, we need \( \frac{17}{2} \) O₂ on the left side.
\[ C_6H_{10} + \frac{17}{2} O_2 \rightarrow 6CO_2 + 5H_2O \]
5. **Multiply through by 2 to eliminate the fraction:**
\[ 2C_6H_{10} + 17O_2 \rightarrow 12CO_2 + 10H_2O \]
### (iv) Combustion of Toluene (C₇H₈)
1. **Write the unbalanced equation:**
\[ C_7H_8 + O_2 \rightarrow CO_2 + H_2O \]
2. **Balance the carbon atoms:**
There are 7 carbon atoms in toluene, so we need 7 CO₂.
\[ C_7H_8 + O_2 \rightarrow 7CO_2 + H_2O \]
3. **Balance the hydrogen atoms:**
There are 8 hydrogen atoms in toluene, so we need 4 H₂O.
\[ C_7H_8 + O_2 \rightarrow 7CO_2 + 4H_2O \]
4. **Balance the oxygen atoms:**
On the right side, we have \(7 \times 2 = 14\) from CO₂ and \(4 \times 1 = 4\) from H₂O, totaling 18 oxygen atoms. Therefore, we need \(9 O₂\) on the left side.
\[ C_7H_8 + 9O_2 \rightarrow 7CO_2 + 4H_2O \]
### Final Balanced Equations
1. **Butane:**
\[ 2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O \]
2. **Pentene:**
\[ 2C_5H_{10} + 15O_2 \rightarrow 10CO_2 + 10H_2O \]
3. **Hexyne:**
\[ 2C_6H_{10} + 17O_2 \rightarrow 12CO_2 + 10H_2O \]
4. **Toluene:**
\[ C_7H_8 + 9O_2 \rightarrow 7CO_2 + 4H_2O \]
