For the following reaction, 5.87 grams of oxygen gas are mixed with excess hydrochloric acid. The reaction yields 4.91 grams of water. 4HCI (aq) + O2 (g) - 2H20 (1) + 2C12 (g) (1) What is the theoretical yield of water? grams (2) What is the percent yield for this reaction? %
For the following reaction, 5.87 grams of oxygen gas are mixed with excess hydrochloric acid. The reaction yields 4.91 grams of water. 4HCI (aq) + O2 (g) - 2H20 (1) + 2C12 (g) (1) What is the theoretical yield of water? grams (2) What is the percent yield for this reaction? %
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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![**Understanding Stoichiometry: Chemical Reactions and Yield Calculations**
**Problem Statement:**
For the following reaction, **5.87 grams of oxygen gas** are mixed with excess **hydrochloric acid**. The reaction yields **4.91 grams of water**.
\[ 4HCl \, \text{(aq)} + O_2 \, \text{(g)} \longrightarrow 2H_2O \, \text{(l)} + 2Cl_2 \, \text{(g)} \]
1. **What is the theoretical yield of water?** _____ grams
2. **What is the percent yield for this reaction?** _____ %
Let's break down the problem to find our answers:
1. **Calculating the Theoretical Yield of Water:**
First, we need to calculate the theoretical yield of water based on the given amount of oxygen gas and the stoichiometry of the reaction.
2. **Calculating the Percent Yield:**
Once we have the theoretical yield, we can use it to determine the percent yield using the actual yield provided in the problem statement.
**Steps to Calculate the Theoretical Yield:**
1. **Balanced Chemical Equation:**
\[ 4HCl \, \text{(aq)} + O_2 \, \text{(g)} \longrightarrow 2H_2O \, \text{(l)} + 2Cl_2 \, \text{(g)} \]
2. **Molar Masses Calculation:**
- Molar mass of \( O_2 \) (Oxygen gas) = \( 2 \times 16.00 \, \text{g/mol} = 32.00 \, \text{g/mol} \)
- Molar mass of \( H_2O \) (Water) = \( 2 \times 1.01 \, \text{g/mol} + 16.00 \, \text{g/mol} = 18.02 \, \text{g/mol} \)
3. **Convert grams of \( O_2 \) to moles:**
\[ \text{Moles of } O_2 = \frac{5.87 \, \text{g}}{32.00 \, \text{g/mol}} = 0.1834](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa527ffdc-16df-4a1f-9953-96956a75d05e%2Fd9e20354-4c24-42d7-995e-e27c9a1ca7ef%2Fqvt9r3o.png&w=3840&q=75)
Transcribed Image Text:**Understanding Stoichiometry: Chemical Reactions and Yield Calculations**
**Problem Statement:**
For the following reaction, **5.87 grams of oxygen gas** are mixed with excess **hydrochloric acid**. The reaction yields **4.91 grams of water**.
\[ 4HCl \, \text{(aq)} + O_2 \, \text{(g)} \longrightarrow 2H_2O \, \text{(l)} + 2Cl_2 \, \text{(g)} \]
1. **What is the theoretical yield of water?** _____ grams
2. **What is the percent yield for this reaction?** _____ %
Let's break down the problem to find our answers:
1. **Calculating the Theoretical Yield of Water:**
First, we need to calculate the theoretical yield of water based on the given amount of oxygen gas and the stoichiometry of the reaction.
2. **Calculating the Percent Yield:**
Once we have the theoretical yield, we can use it to determine the percent yield using the actual yield provided in the problem statement.
**Steps to Calculate the Theoretical Yield:**
1. **Balanced Chemical Equation:**
\[ 4HCl \, \text{(aq)} + O_2 \, \text{(g)} \longrightarrow 2H_2O \, \text{(l)} + 2Cl_2 \, \text{(g)} \]
2. **Molar Masses Calculation:**
- Molar mass of \( O_2 \) (Oxygen gas) = \( 2 \times 16.00 \, \text{g/mol} = 32.00 \, \text{g/mol} \)
- Molar mass of \( H_2O \) (Water) = \( 2 \times 1.01 \, \text{g/mol} + 16.00 \, \text{g/mol} = 18.02 \, \text{g/mol} \)
3. **Convert grams of \( O_2 \) to moles:**
\[ \text{Moles of } O_2 = \frac{5.87 \, \text{g}}{32.00 \, \text{g/mol}} = 0.1834
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