Answer :
Let's solve this problem step-by-step by analyzing the chemical reaction of methane combustion for each set of initial reactant amounts.
The chemical equation for the combustion of methane is:
[tex]\[ \text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \][/tex]
This means 1 molecule of methane (CH₄) reacts with 2 molecules of oxygen (O₂) to produce 1 molecule of carbon dioxide (CO₂) and 2 molecules of water (H₂O).
### 1. For Reactants: 1 CH₄ and 5 O₂
- Products Formed:
- 1 CH₄ can react with 2 O₂ to produce 1 CO₂ and 2 H₂O.
- Leftovers:
- CH₄ is completely used up (0 CH₄ leftover).
- Initially, 5 O₂ was present. After reaction, [tex]\(5 - 2 = 3\)[/tex] O₂ molecules are left over.
- Limiting Reactant: CH₄
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|c|} \hline \multicolumn{2}{|c|}{Reactants} & \multicolumn{2}{|c|}{Products} & \multicolumn{2}{|c|}{Leftovers} & Limiting Reactant \\ \hline CH_4 & O_2 & CO_2 & H_2O & CH_4 & O_2 & \\ \hline 1 & 5 & 1 & 2 & 0 & 3 & CH_4 \\ \hline \end{tabular} \][/tex]
### 2. For Reactants: 4 CH₄ and 6 O₂
- Products Formed:
- To react completely with 4 CH₄, we need [tex]\(4 \times 2 = 8\)[/tex] O₂ molecules, but only 6 O₂ are available.
- Therefore, only [tex]\(6 \div 2 = 3\)[/tex] CH₄ and [tex]\(3 \times 2 = 6\)[/tex] O₂ will react to produce [tex]\(3 \times 1 = 3\)[/tex] CO₂ and [tex]\(3 \times 2 = 6\)[/tex] H₂O.
- Leftovers:
- CH₄ leftover: [tex]\(4 - 3 = 1\)[/tex]
- O₂ leftover: [tex]\(6 - 6 = 0\)[/tex]
- Limiting Reactant: O₂
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|c|} \hline \multicolumn{2}{|c|}{Reactants} & \multicolumn{2}{|c|}{Products} & \multicolumn{2}{|c|}{Leftovers} & Limiting Reactant \\ \hline CH_4 & O_2 & CO_2 & H_2O & CH_4 & O_2 & \\ \hline 4 & 6 & 3 & 6 & 1 & 0 & O_2 \\ \hline \end{tabular} \][/tex]
### 3. For Reactants: 5 CH₄ and 4 O₂
- Products Formed:
- To react completely with 5 CH₄, we need [tex]\(5 \times 2 = 10\)[/tex] O₂ molecules, but only 4 O₂ are available.
- Therefore, only [tex]\(4 \div 2 = 2\)[/tex] CH₄ and [tex]\(2 \times 2 = 4\)[/tex] O₂ will react to produce [tex]\(2 \times 1 = 2\)[/tex] CO₂ and [tex]\(2 \times 2 = 4\)[/tex] H₂O.
- Leftovers:
- CH₄ leftover: [tex]\(5 - 2 = 3\)[/tex]
- O₂ leftover: [tex]\(4 - 4 = 0\)[/tex]
- Limiting Reactant: O₂
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|c|} \hline \multicolumn{2}{|c|}{Reactants} & \multicolumn{2}{|c|}{Products} & \multicolumn{2}{|c|}{Leftovers} & Limiting Reactant \\ \hline CH_4 & O_2 & CO_2 & H_2O & CH_4 & O_2 & \\ \hline 5 & 4 & 2 & 4 & 3 & 0 & O_2 \\ \hline \end{tabular} \][/tex]
### 4. For Reactants: 3 CH₄ and 8 O₂
- Products Formed:
- 3 CH₄ can react with [tex]\(3 \times 2 = 6\)[/tex] O₂ to produce [tex]\(3 \times 1 = 3\)[/tex] CO₂ and [tex]\(3 \times 2 = 6\)[/tex] H₂O.
- Leftovers:
- CH₄ is completely used up (0 CH₄ leftover).
- Initially, 8 O₂ was present. After reaction, [tex]\(8 - 6 = 2\)[/tex] O₂ molecules are left over.
- Limiting Reactant: CH₄
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|c|} \hline \multicolumn{2}{|c|}{Reactants} & \multicolumn{2}{|c|}{Products} & \multicolumn{2}{|c|}{Leftovers} & Limiting Reactant \\ \hline CH_4 & O_2 & CO_2 & H_2O & CH_4 & O_2 & \\ \hline 3 & 8 & 3 & 6 & 0 & 2 & CH_4 \\ \hline \end{tabular} \][/tex]
The chemical equation for the combustion of methane is:
[tex]\[ \text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \][/tex]
This means 1 molecule of methane (CH₄) reacts with 2 molecules of oxygen (O₂) to produce 1 molecule of carbon dioxide (CO₂) and 2 molecules of water (H₂O).
### 1. For Reactants: 1 CH₄ and 5 O₂
- Products Formed:
- 1 CH₄ can react with 2 O₂ to produce 1 CO₂ and 2 H₂O.
- Leftovers:
- CH₄ is completely used up (0 CH₄ leftover).
- Initially, 5 O₂ was present. After reaction, [tex]\(5 - 2 = 3\)[/tex] O₂ molecules are left over.
- Limiting Reactant: CH₄
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|c|} \hline \multicolumn{2}{|c|}{Reactants} & \multicolumn{2}{|c|}{Products} & \multicolumn{2}{|c|}{Leftovers} & Limiting Reactant \\ \hline CH_4 & O_2 & CO_2 & H_2O & CH_4 & O_2 & \\ \hline 1 & 5 & 1 & 2 & 0 & 3 & CH_4 \\ \hline \end{tabular} \][/tex]
### 2. For Reactants: 4 CH₄ and 6 O₂
- Products Formed:
- To react completely with 4 CH₄, we need [tex]\(4 \times 2 = 8\)[/tex] O₂ molecules, but only 6 O₂ are available.
- Therefore, only [tex]\(6 \div 2 = 3\)[/tex] CH₄ and [tex]\(3 \times 2 = 6\)[/tex] O₂ will react to produce [tex]\(3 \times 1 = 3\)[/tex] CO₂ and [tex]\(3 \times 2 = 6\)[/tex] H₂O.
- Leftovers:
- CH₄ leftover: [tex]\(4 - 3 = 1\)[/tex]
- O₂ leftover: [tex]\(6 - 6 = 0\)[/tex]
- Limiting Reactant: O₂
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|c|} \hline \multicolumn{2}{|c|}{Reactants} & \multicolumn{2}{|c|}{Products} & \multicolumn{2}{|c|}{Leftovers} & Limiting Reactant \\ \hline CH_4 & O_2 & CO_2 & H_2O & CH_4 & O_2 & \\ \hline 4 & 6 & 3 & 6 & 1 & 0 & O_2 \\ \hline \end{tabular} \][/tex]
### 3. For Reactants: 5 CH₄ and 4 O₂
- Products Formed:
- To react completely with 5 CH₄, we need [tex]\(5 \times 2 = 10\)[/tex] O₂ molecules, but only 4 O₂ are available.
- Therefore, only [tex]\(4 \div 2 = 2\)[/tex] CH₄ and [tex]\(2 \times 2 = 4\)[/tex] O₂ will react to produce [tex]\(2 \times 1 = 2\)[/tex] CO₂ and [tex]\(2 \times 2 = 4\)[/tex] H₂O.
- Leftovers:
- CH₄ leftover: [tex]\(5 - 2 = 3\)[/tex]
- O₂ leftover: [tex]\(4 - 4 = 0\)[/tex]
- Limiting Reactant: O₂
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|c|} \hline \multicolumn{2}{|c|}{Reactants} & \multicolumn{2}{|c|}{Products} & \multicolumn{2}{|c|}{Leftovers} & Limiting Reactant \\ \hline CH_4 & O_2 & CO_2 & H_2O & CH_4 & O_2 & \\ \hline 5 & 4 & 2 & 4 & 3 & 0 & O_2 \\ \hline \end{tabular} \][/tex]
### 4. For Reactants: 3 CH₄ and 8 O₂
- Products Formed:
- 3 CH₄ can react with [tex]\(3 \times 2 = 6\)[/tex] O₂ to produce [tex]\(3 \times 1 = 3\)[/tex] CO₂ and [tex]\(3 \times 2 = 6\)[/tex] H₂O.
- Leftovers:
- CH₄ is completely used up (0 CH₄ leftover).
- Initially, 8 O₂ was present. After reaction, [tex]\(8 - 6 = 2\)[/tex] O₂ molecules are left over.
- Limiting Reactant: CH₄
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|c|} \hline \multicolumn{2}{|c|}{Reactants} & \multicolumn{2}{|c|}{Products} & \multicolumn{2}{|c|}{Leftovers} & Limiting Reactant \\ \hline CH_4 & O_2 & CO_2 & H_2O & CH_4 & O_2 & \\ \hline 3 & 8 & 3 & 6 & 0 & 2 & CH_4 \\ \hline \end{tabular} \][/tex]
