Question 12 of 25 Submit What mass grams of nitric acid , HNO, is required to neutralize (completely react with) 4.30 |g of Ca(OH)2 according to the acid-base reaction: 2 HNO,(aq) + Ca(OH)2(aq) → 2 H,0(1) + Ca(NO3)2(aq) 0.0580 mol GatNO 4.30
Question 12 of 25 Submit What mass grams of nitric acid , HNO, is required to neutralize (completely react with) 4.30 |g of Ca(OH)2 according to the acid-base reaction: 2 HNO,(aq) + Ca(OH)2(aq) → 2 H,0(1) + Ca(NO3)2(aq) 0.0580 mol GatNO 4.30
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 12 of 25**
**What mass (in grams) of nitric acid, HNO₃, is required to neutralize (completely react with) 4.30 g of Ca(OH)₂ according to the acid-base reaction:**
\[ 2 \, \text{HNO}_3(aq) + \text{Ca(OH)}_2(aq) \rightarrow 2 \, \text{H}_2\text{O}(l) + \text{Ca(NO}_3\text{)}_2(aq) \]
**Starting Calculation:**
- **Starting Amount:** 4.30 g \( \text{Ca(OH)}_2 \)
- **Conversion Factor:**
\(\frac{0.0580 \, \text{mol} \, \text{Ca(NO}_3\text{)}_2}{74.10 \, \text{g} \, \text{Ca(OH)}_2}\)
- **Calculation Steps:**
\[
4.30 \, \text{g Ca(OH)}_2 \times \frac{0.0580 \, \text{mol} \, \text{Ca(NO}_3\text{)}_2}{74.10 \, \text{g Ca(OH)}_2} \times \frac{1 \, \text{mol Ca(OH)}_2}{1 \, \text{mol Ca(NO}_3\text{)}_2} \times \frac{2 \, \text{mol HNO}_3}{1 \, \text{mol Ca(OH)}_2} \times \frac{63.02 \, \text{g HNO}_3}{1 \, \text{mol HNO}_3} = 7.31 \, \text{g HNO}_3
\]
**Interface Explanation:**
- **Interactive Elements:**
- **Add Factor:** Allows you to multiply or add factors.
- **Answer Box:** Input the calculated answer.
- **Reset Button:** Clears the inputs for a new attempt.
**Units Menu:**
- Various units and constants are available for selection:
- Conversion factors include grams and moles for \( \text{Ca(OH)}_2 \), \( \text{Ca(NO}_3\text{)}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe60afec2-8270-463c-8466-04e9f64c353b%2F7d5522fe-2bf3-4bb5-ac58-1bf2412c9ff5%2Fbaopwou_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 12 of 25**
**What mass (in grams) of nitric acid, HNO₃, is required to neutralize (completely react with) 4.30 g of Ca(OH)₂ according to the acid-base reaction:**
\[ 2 \, \text{HNO}_3(aq) + \text{Ca(OH)}_2(aq) \rightarrow 2 \, \text{H}_2\text{O}(l) + \text{Ca(NO}_3\text{)}_2(aq) \]
**Starting Calculation:**
- **Starting Amount:** 4.30 g \( \text{Ca(OH)}_2 \)
- **Conversion Factor:**
\(\frac{0.0580 \, \text{mol} \, \text{Ca(NO}_3\text{)}_2}{74.10 \, \text{g} \, \text{Ca(OH)}_2}\)
- **Calculation Steps:**
\[
4.30 \, \text{g Ca(OH)}_2 \times \frac{0.0580 \, \text{mol} \, \text{Ca(NO}_3\text{)}_2}{74.10 \, \text{g Ca(OH)}_2} \times \frac{1 \, \text{mol Ca(OH)}_2}{1 \, \text{mol Ca(NO}_3\text{)}_2} \times \frac{2 \, \text{mol HNO}_3}{1 \, \text{mol Ca(OH)}_2} \times \frac{63.02 \, \text{g HNO}_3}{1 \, \text{mol HNO}_3} = 7.31 \, \text{g HNO}_3
\]
**Interface Explanation:**
- **Interactive Elements:**
- **Add Factor:** Allows you to multiply or add factors.
- **Answer Box:** Input the calculated answer.
- **Reset Button:** Clears the inputs for a new attempt.
**Units Menu:**
- Various units and constants are available for selection:
- Conversion factors include grams and moles for \( \text{Ca(OH)}_2 \), \( \text{Ca(NO}_3\text{)}
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