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Sagot :
To solve this problem, follow these steps:
### Step 1: Understand the Balanced Chemical Equation
The balanced chemical equation provided is:
[tex]\[ 2 \text{C}_2\text{H}_2(g) + 5 \text{O}_2(g) \rightarrow 4 \text{CO}_2(g) + 2 \text{H}_2\text{O}(g) \][/tex]
### Step 2: Molar Ratio from the Balanced Equation
From the balanced equation, you can see that 2 moles of [tex]\(\text{C}_2\text{H}_2\)[/tex] react with 5 moles of [tex]\(\text{O}_2\)[/tex].
### Step 3: Determine the Molar Ratio
The molar ratio of [tex]\(\text{C}_2\text{H}_2\)[/tex] to [tex]\(\text{O}_2\)[/tex] is:
[tex]\[ \frac{2 \text{ mol } \text{C}_2\text{H}_2}{5 \text{ mol }\text{O}_2} \][/tex]
### Step 4: Calculate the Moles of [tex]\(\text{C}_2\text{H}_2\)[/tex]
You are given 12.0 moles of [tex]\(\text{O}_2\)[/tex]. To find the moles of [tex]\(\text{C}_2\text{H}_2\)[/tex] that react with this amount of [tex]\(\text{O}_2\)[/tex], use the molar ratio:
[tex]\[ \text{moles of }\text{C}_2\text{H}_2 = 12.0 \text{ mol }\text{O}_2 \times \frac{2 \text{ mol }\text{C}_2\text{H}_2}{5 \text{ mol }\text{O}_2} \][/tex]
[tex]\[ \text{moles of }\text{C}_2\text{H}_2 = 12.0 \times \frac{2}{5} = 4.8 \text{ mol }\text{C}_2\text{H}_2 \][/tex]
### Step 5: Volume of [tex]\(\text{C}_2\text{H}_2\)[/tex] at STP
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, to find the volume of 4.8 moles of [tex]\(\text{C}_2\text{H}_2\)[/tex], multiply the moles by the volume occupied by 1 mole:
[tex]\[ \text{Volume} = 4.8 \text{ mol }\text{C}_2\text{H}_2 \times 22.4 \text{ L/mol} \][/tex]
[tex]\[ \text{Volume} = 107.52 \text{ liters} \][/tex]
### Conclusion
Thus, 4.8 moles of [tex]\(\text{C}_2\text{H}_2\)[/tex] will react with 12.0 moles of [tex]\(\text{O}_2\)[/tex], and the volume of [tex]\(\text{C}_2\text{H}_2\)[/tex] at STP required for this reaction is 107.52 liters.
Your dimensional setup for the two-step calculation should look like:
[tex]\[ 12.0 \text{ mol }\text{O}_2 \times \frac{2 \text{ mol }\text{C}_2\text{H}_2}{5 \text{ mol }\text{O}_2} = 4.8 \text{ mol }\text{C}_2\text{H}_2 \quad \Rightarrow \quad 4.8 \text{ mol }\text{C}_2\text{H}_2 \times 22.4 \text{ L/mol} = 107.52 \text{ liters} \][/tex]
### Step 1: Understand the Balanced Chemical Equation
The balanced chemical equation provided is:
[tex]\[ 2 \text{C}_2\text{H}_2(g) + 5 \text{O}_2(g) \rightarrow 4 \text{CO}_2(g) + 2 \text{H}_2\text{O}(g) \][/tex]
### Step 2: Molar Ratio from the Balanced Equation
From the balanced equation, you can see that 2 moles of [tex]\(\text{C}_2\text{H}_2\)[/tex] react with 5 moles of [tex]\(\text{O}_2\)[/tex].
### Step 3: Determine the Molar Ratio
The molar ratio of [tex]\(\text{C}_2\text{H}_2\)[/tex] to [tex]\(\text{O}_2\)[/tex] is:
[tex]\[ \frac{2 \text{ mol } \text{C}_2\text{H}_2}{5 \text{ mol }\text{O}_2} \][/tex]
### Step 4: Calculate the Moles of [tex]\(\text{C}_2\text{H}_2\)[/tex]
You are given 12.0 moles of [tex]\(\text{O}_2\)[/tex]. To find the moles of [tex]\(\text{C}_2\text{H}_2\)[/tex] that react with this amount of [tex]\(\text{O}_2\)[/tex], use the molar ratio:
[tex]\[ \text{moles of }\text{C}_2\text{H}_2 = 12.0 \text{ mol }\text{O}_2 \times \frac{2 \text{ mol }\text{C}_2\text{H}_2}{5 \text{ mol }\text{O}_2} \][/tex]
[tex]\[ \text{moles of }\text{C}_2\text{H}_2 = 12.0 \times \frac{2}{5} = 4.8 \text{ mol }\text{C}_2\text{H}_2 \][/tex]
### Step 5: Volume of [tex]\(\text{C}_2\text{H}_2\)[/tex] at STP
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, to find the volume of 4.8 moles of [tex]\(\text{C}_2\text{H}_2\)[/tex], multiply the moles by the volume occupied by 1 mole:
[tex]\[ \text{Volume} = 4.8 \text{ mol }\text{C}_2\text{H}_2 \times 22.4 \text{ L/mol} \][/tex]
[tex]\[ \text{Volume} = 107.52 \text{ liters} \][/tex]
### Conclusion
Thus, 4.8 moles of [tex]\(\text{C}_2\text{H}_2\)[/tex] will react with 12.0 moles of [tex]\(\text{O}_2\)[/tex], and the volume of [tex]\(\text{C}_2\text{H}_2\)[/tex] at STP required for this reaction is 107.52 liters.
Your dimensional setup for the two-step calculation should look like:
[tex]\[ 12.0 \text{ mol }\text{O}_2 \times \frac{2 \text{ mol }\text{C}_2\text{H}_2}{5 \text{ mol }\text{O}_2} = 4.8 \text{ mol }\text{C}_2\text{H}_2 \quad \Rightarrow \quad 4.8 \text{ mol }\text{C}_2\text{H}_2 \times 22.4 \text{ L/mol} = 107.52 \text{ liters} \][/tex]
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