A mixture of 25.0 g of NO and an excess of O2 reacts according to the balanced equation _2_NO(g) + ___O2(g) --> _2_NO2(g) How many liters of NO2, at STP, are produced?
A mixture of 25.0 g of NO and an excess of O2 reacts according to the balanced equation _2_NO(g) + ___O2(g) --> _2_NO2(g) How many liters of NO2, at STP, are produced?
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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- A mixture of 25.0 g of NO and an excess of O2 reacts according to the balanced equation
_2_NO(g) + ___O2(g) --> _2_NO2(g)
How many liters of NO2, at STP, are produced?
![**Problem Statement:**
A mixture of 25.0 g of NO and an excess of O₂ reacts according to the balanced equation:
\[ 2 \text{NO(g)} + \_\_\_ \, \text{O}_2\text{(g)} \rightarrow 2 \text{NO}_2\text{(g)} \]
How many liters of NO₂, at STP, are produced?
---
**Balanced Chemical Equation:**
The equation provided is partially balanced. The correct balanced equation is:
\[ 2 \text{NO(g)} + 1 \, \text{O}_2\text{(g)} \rightarrow 2 \text{NO}_2\text{(g)} \]
**Explanation:**
- Two moles of Nitric Oxide (NO) gas react with one mole of Oxygen (O₂) gas to produce two moles of Nitrogen Dioxide (NO₂) gas.
**Stoichiometry Explanation:**
- Determine the moles of NO:
NO has a molar mass of approximately 30.01 g/mol.
Moles of NO = \(\frac{25.0 \, \text{g}}{30.01 \, \text{g/mol}} \approx 0.833 \, \text{moles}\)
- According to the balanced equation, 2 moles of NO produce 2 moles of NO₂.
- Since the reaction is at STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 liters.
- Liters of NO₂ = 0.833 moles × 22.4 L/mol = 18.66 liters
Thus, 18.66 liters of NO₂ are produced at STP.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbcea870b-a558-4e25-bd85-f50b869e1110%2Fb1c3626a-661f-4bc0-a30a-af5a6cbc82de%2F7totvg4_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A mixture of 25.0 g of NO and an excess of O₂ reacts according to the balanced equation:
\[ 2 \text{NO(g)} + \_\_\_ \, \text{O}_2\text{(g)} \rightarrow 2 \text{NO}_2\text{(g)} \]
How many liters of NO₂, at STP, are produced?
---
**Balanced Chemical Equation:**
The equation provided is partially balanced. The correct balanced equation is:
\[ 2 \text{NO(g)} + 1 \, \text{O}_2\text{(g)} \rightarrow 2 \text{NO}_2\text{(g)} \]
**Explanation:**
- Two moles of Nitric Oxide (NO) gas react with one mole of Oxygen (O₂) gas to produce two moles of Nitrogen Dioxide (NO₂) gas.
**Stoichiometry Explanation:**
- Determine the moles of NO:
NO has a molar mass of approximately 30.01 g/mol.
Moles of NO = \(\frac{25.0 \, \text{g}}{30.01 \, \text{g/mol}} \approx 0.833 \, \text{moles}\)
- According to the balanced equation, 2 moles of NO produce 2 moles of NO₂.
- Since the reaction is at STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 liters.
- Liters of NO₂ = 0.833 moles × 22.4 L/mol = 18.66 liters
Thus, 18.66 liters of NO₂ are produced at STP.
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