What volume of water vapor gas would be produced from the combustion of 570.89 grams of propane (C3H8) with 2,808.904 grams of oxygen gas, under a pressure of 0.96 atm and a temperature of 350. degrees C? Given: C3H8(g) + 5 O2(g)- 3 CO2(g) + 4 H20(g) ---> (OR C3 Hg ("g") + 5 O2 ("g") right arrow 3 C O2 ("g") + 4 H20 ("g") R = 0.08206 Latm mol K Do not type units with your answer Your Answer:
What volume of water vapor gas would be produced from the combustion of 570.89 grams of propane (C3H8) with 2,808.904 grams of oxygen gas, under a pressure of 0.96 atm and a temperature of 350. degrees C? Given: C3H8(g) + 5 O2(g)- 3 CO2(g) + 4 H20(g) ---> (OR C3 Hg ("g") + 5 O2 ("g") right arrow 3 C O2 ("g") + 4 H20 ("g") R = 0.08206 Latm mol K Do not type units with your answer Your Answer:
Chemistry: Matter and Change
1st Edition
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Chapter13: Gases
Section: Chapter Questions
Problem 88A
Related questions
Question
![**Combustion of Propane and Water Vapor Production**
**Problem Statement:**
What volume of water vapor gas would be produced from the combustion of 570.89 grams of propane (C₃H₈) with 2,808.904 grams of oxygen gas, under a pressure of 0.96 atm and a temperature of 350 degrees Celsius?
**Chemical Reaction:**
C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)
Alternatively written:
\( \text{C}_3\text{H}_8 \text{(g)} + 5 \, \text{O}_2 \text{(g)} \rightarrow 3 \, \text{CO}_2 \text{(g)} + 4 \, \text{H}_2\text{O} \text{(g)} \)
**Gas Constant:**
\[ R = 0.08206 \, \text{L atm/mol K} \]
**Instructions:**
Please do not include units with your answer.
**Response Section:**
Your Answer: _______](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F80bd6999-b702-4bab-9a9d-3d78c3747c19%2Fe147e9cb-540f-4e69-a643-104a348f9cbe%2Fbrv1rgd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Combustion of Propane and Water Vapor Production**
**Problem Statement:**
What volume of water vapor gas would be produced from the combustion of 570.89 grams of propane (C₃H₈) with 2,808.904 grams of oxygen gas, under a pressure of 0.96 atm and a temperature of 350 degrees Celsius?
**Chemical Reaction:**
C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)
Alternatively written:
\( \text{C}_3\text{H}_8 \text{(g)} + 5 \, \text{O}_2 \text{(g)} \rightarrow 3 \, \text{CO}_2 \text{(g)} + 4 \, \text{H}_2\text{O} \text{(g)} \)
**Gas Constant:**
\[ R = 0.08206 \, \text{L atm/mol K} \]
**Instructions:**
Please do not include units with your answer.
**Response Section:**
Your Answer: _______
Expert Solution
Step 1
Limiting reactant :-
A reactant which is present in lesser number of moles than required as per balanced chemical equation is known as limiting reactant
The amount if product formed will depend on limiting reactant
Step by step
Solved in 4 steps
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