What volume in liters of fluorine gas is needed to form 999 L of sulfur hexafluoride gas if the following reaction takes place at 2.00 atm and 273.15 K: S(s) + 3F2(g) → SF, (g)?
What volume in liters of fluorine gas is needed to form 999 L of sulfur hexafluoride gas if the following reaction takes place at 2.00 atm and 273.15 K: S(s) + 3F2(g) → SF, (g)?
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Problem Statement:**
What volume in liters of fluorine gas is needed to form 999 L of sulfur hexafluoride gas if the following reaction takes place at 2.00 atm and 273.15 K:
\[ \text{S(s)} + 3\text{F}_2\text{(g)} \rightarrow \text{SF}_6\text{(g)} \]
**Explanation:**
Given the reaction:
\[ \text{S(s)} + 3\text{F}_2\text{(g)} \rightarrow \text{SF}_6\text{(g)} \]
This stoichiometric equation tells us that one mole of sulfur reacts with three moles of fluorine gas to produce one mole of sulfur hexafluoride. According to the Ideal Gas Law, under the same conditions of temperature and pressure, gases react in volumes that are in the same ratio as their coefficients in the balanced equation. Thus, to produce one volume of sulfur hexafluoride, three volumes of fluorine gas are required.
For 999 L of sulfur hexafluoride:
- You need 3 times that volume of fluorine gas:
\[ 999 \, \text{L SF}_6 \times 3 = 2997 \, \text{L of } \text{F}_2\]
**Calculation:**
Therefore, 2997 liters of fluorine gas are required to produce 999 liters of sulfur hexafluoride at 2.00 atm and 273.15 K.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F18d66836-1cf9-4957-9659-b615a4dc4665%2F6b30682f-3751-425d-87e9-f0ddf31d46b3%2F5h0x1nn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
What volume in liters of fluorine gas is needed to form 999 L of sulfur hexafluoride gas if the following reaction takes place at 2.00 atm and 273.15 K:
\[ \text{S(s)} + 3\text{F}_2\text{(g)} \rightarrow \text{SF}_6\text{(g)} \]
**Explanation:**
Given the reaction:
\[ \text{S(s)} + 3\text{F}_2\text{(g)} \rightarrow \text{SF}_6\text{(g)} \]
This stoichiometric equation tells us that one mole of sulfur reacts with three moles of fluorine gas to produce one mole of sulfur hexafluoride. According to the Ideal Gas Law, under the same conditions of temperature and pressure, gases react in volumes that are in the same ratio as their coefficients in the balanced equation. Thus, to produce one volume of sulfur hexafluoride, three volumes of fluorine gas are required.
For 999 L of sulfur hexafluoride:
- You need 3 times that volume of fluorine gas:
\[ 999 \, \text{L SF}_6 \times 3 = 2997 \, \text{L of } \text{F}_2\]
**Calculation:**
Therefore, 2997 liters of fluorine gas are required to produce 999 liters of sulfur hexafluoride at 2.00 atm and 273.15 K.
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