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

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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{)}
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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