Question 16 of 24 Submit How many grams of oxygen gas are produced when 2.43 x 10-ªg of KCIo, are completely reacted according to the following chemical equation: 2 KCIO (s) → 2 KCI(s) + 3 02(g) STARTING AMOUNT ADD FACTOR ANSWER RESET

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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**Question 16 of 24**

**Chemical Reaction Problem:**

How many grams of oxygen gas are produced when \(2.43 \times 10^{-4}\) g of KClO\(_3\) are completely reacted according to the following chemical equation:

\[ 2 \text{ KClO}_3(s) \rightarrow 2 \text{ KCl}(s) + 3 \text{ O}_2(g) \]

**Calculation Interface:**

- **Starting Amount Box:** This section requires input for the starting amount of the substance in question.
- **Equation Structure:** The interface employs multiplication and division to convert between units.

**Factors Available for Calculation:**

- \(1\)
- \(32.00\)
- \(1.90 \times 10^{-4}\)
- \(74.55\)
- \(1.98 \times 10^{-6}\)
- \(2.43 \times 10^{-4}\)
- \(2.97 \times 10^{-6}\)

**Conversion and Calculation Options:**

- **Grams to Moles and Vice Versa:**
  - \(9.52 \times 10^{-5}\) g KClO\(_3\)
  - \(2\) mol KClO\(_3\)
  - \(3\) mol KCl
  - \(5.95 \times 10^{-6}\) mol O\(_2\)
  - \(6.022 \times 10^{23}\) g O\(_2\)
  - \(122.55\) g KCl

**Functional Buttons:**

- **Add Factor:** Allows for the inclusion of an additional conversion or reaction factor.
- **Answer Box:** Entry for the calculated result in the desired unit.
- **Reset Button:** Clears current calculations to restart the problem.

**Resource Note:**

Access additional resources at the bottom of the interface for further exploration and learning.
Transcribed Image Text:**Question 16 of 24** **Chemical Reaction Problem:** How many grams of oxygen gas are produced when \(2.43 \times 10^{-4}\) g of KClO\(_3\) are completely reacted according to the following chemical equation: \[ 2 \text{ KClO}_3(s) \rightarrow 2 \text{ KCl}(s) + 3 \text{ O}_2(g) \] **Calculation Interface:** - **Starting Amount Box:** This section requires input for the starting amount of the substance in question. - **Equation Structure:** The interface employs multiplication and division to convert between units. **Factors Available for Calculation:** - \(1\) - \(32.00\) - \(1.90 \times 10^{-4}\) - \(74.55\) - \(1.98 \times 10^{-6}\) - \(2.43 \times 10^{-4}\) - \(2.97 \times 10^{-6}\) **Conversion and Calculation Options:** - **Grams to Moles and Vice Versa:** - \(9.52 \times 10^{-5}\) g KClO\(_3\) - \(2\) mol KClO\(_3\) - \(3\) mol KCl - \(5.95 \times 10^{-6}\) mol O\(_2\) - \(6.022 \times 10^{23}\) g O\(_2\) - \(122.55\) g KCl **Functional Buttons:** - **Add Factor:** Allows for the inclusion of an additional conversion or reaction factor. - **Answer Box:** Entry for the calculated result in the desired unit. - **Reset Button:** Clears current calculations to restart the problem. **Resource Note:** Access additional resources at the bottom of the interface for further exploration and learning.
**Question 17 of 24**

**Determine the number of molecules of Cr₂O₃ that form when 1.34 x 10³ g of oxygen completely reacts according to the following equation:**

\[ 4 \text{ Cr(s)} + 3 \text{ O}_2\text{(g)} \rightarrow 2 \text{ Cr}_2\text{O}_3\text{(s)} \]

---

**Interactive Calculation Interface:**

- **Starting Amount:** Two boxes for input values.
- **Equation Setup:** 
  - Multiplication setup with boxes for input, multiplication, and division.
  
**Controls:**

- **Add Factor:** Enter additional numerical factors.
- **Answer:** Display box for the calculated answer.
- **Reset:** Button to clear inputs.

---

**Available Inputs:**

- **Numerical Values:**
  - \( 2.52 \times 10^{25} \)
  - \( 3.78 \times 10^{25} \)
  - \( 2 \)
  - \( 1 \)
  - \( 1.34 \times 10^{3} \)
  - \( 152.00 \)
  - \( 52.00 \)
  
- **Constants and Conversions:**
  - \( 6.022 \times 10^{23} \) (Avogadro's number)
  - \( 4 \)
  - \( 1.68 \times 10^{25} \)
  - \( 32.00 \) (possibly molar mass of O₂)
  - \( 3 \)

- **Units for Inputs:**
  - g Cr
  - g Cr₂O₃
  - g O₂
  - mol Cr₂O₃
  - mol Cr
  - molecules Cr₂O₃
  - mol O₂

---

**Instructions:**

1. Input the starting amount of oxygen in grams.
2. Use the provided constants and conversion factors to calculate the number of molecules of Cr₂O₃ formed.
3. Check your answer using the equation setup and clear any inputs using the reset button if necessary.

**Note:** Tap or pull up for additional resources.
Transcribed Image Text:**Question 17 of 24** **Determine the number of molecules of Cr₂O₃ that form when 1.34 x 10³ g of oxygen completely reacts according to the following equation:** \[ 4 \text{ Cr(s)} + 3 \text{ O}_2\text{(g)} \rightarrow 2 \text{ Cr}_2\text{O}_3\text{(s)} \] --- **Interactive Calculation Interface:** - **Starting Amount:** Two boxes for input values. - **Equation Setup:** - Multiplication setup with boxes for input, multiplication, and division. **Controls:** - **Add Factor:** Enter additional numerical factors. - **Answer:** Display box for the calculated answer. - **Reset:** Button to clear inputs. --- **Available Inputs:** - **Numerical Values:** - \( 2.52 \times 10^{25} \) - \( 3.78 \times 10^{25} \) - \( 2 \) - \( 1 \) - \( 1.34 \times 10^{3} \) - \( 152.00 \) - \( 52.00 \) - **Constants and Conversions:** - \( 6.022 \times 10^{23} \) (Avogadro's number) - \( 4 \) - \( 1.68 \times 10^{25} \) - \( 32.00 \) (possibly molar mass of O₂) - \( 3 \) - **Units for Inputs:** - g Cr - g Cr₂O₃ - g O₂ - mol Cr₂O₃ - mol Cr - molecules Cr₂O₃ - mol O₂ --- **Instructions:** 1. Input the starting amount of oxygen in grams. 2. Use the provided constants and conversion factors to calculate the number of molecules of Cr₂O₃ formed. 3. Check your answer using the equation setup and clear any inputs using the reset button if necessary. **Note:** Tap or pull up for additional resources.
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