Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
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]
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.