2.0kg H20. 54. In an accident, a solution containing 2.5 kg of nitric acid was spilled. Two kilograms of Na2CO3 was quickly spread on the area andCO2)was released by the reaction. Was sufficient Na2CO3 used to neutralize all of the acid? Solution The reaction is Na,CO, +2HNO, 2Na + 2NO,) + 2CO, + H,0 Calculate the mass of Na2CO3 required for complete reaction of nitric acid using the process → mol of Na,CO, mass of acid → mol of acid → mass of Na,CO, 1 mel-HNO, 1 mel Na,CO, 106.0 g -1 mass of Na,CO, 2,5 kg)x 1000 g kg x %3D 63.0 g 2 met HNO, mel Na,CO, = 2.1 x 10 g= 2.1 kg No, 2.1 kg is required.
2.0kg H20. 54. In an accident, a solution containing 2.5 kg of nitric acid was spilled. Two kilograms of Na2CO3 was quickly spread on the area andCO2)was released by the reaction. Was sufficient Na2CO3 used to neutralize all of the acid? Solution The reaction is Na,CO, +2HNO, 2Na + 2NO,) + 2CO, + H,0 Calculate the mass of Na2CO3 required for complete reaction of nitric acid using the process → mol of Na,CO, mass of acid → mol of acid → mass of Na,CO, 1 mel-HNO, 1 mel Na,CO, 106.0 g -1 mass of Na,CO, 2,5 kg)x 1000 g kg x %3D 63.0 g 2 met HNO, mel Na,CO, = 2.1 x 10 g= 2.1 kg No, 2.1 kg is required.
Chemistry
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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
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Can anyone step by step （in details）explain how we get 2HNO2 from nitric acid？how does we get the equation？
![**Problem 54: Neutralization of Nitric Acid Spill**
In an accident, a solution containing 2.5 kg of nitric acid (HNO₃) was spilled. Two kilograms of sodium carbonate (Na₂CO₃) was quickly spread on the area, and carbon dioxide (CO₂) was released by the reaction. Was sufficient Na₂CO₃ used to neutralize all of the acid?
**Solution:**
The chemical reaction is:
\[ \text{Na}_2\text{CO}_3 + 2\text{HNO}_3 \rightarrow 2\text{Na}^+ + 2\text{NO}_3^- + 2\text{CO}_2 + \text{H}_2\text{O} \]
To calculate the mass of Na₂CO₃ required for the complete reaction with nitric acid, follow these steps:
1. **Mass of Acid → Moles of Acid → Moles of Na₂CO₃ → Mass of Na₂CO₃**
- Calculate the number of moles of HNO₃:
\[
\frac{2.5 \, \text{kg} \times 1000 \, \text{g/kg}}{63.0 \, \text{g/mol}} = \text{moles of } \text{HNO}_3
\]
- Use the stoichiometry of the balanced reaction:
\[
\text{1 mol Na}_2\text{CO}_3 \text{ is required per 2 mol HNO}_3
\]
- Convert moles of Na₂CO₃ to mass:
\[
\times \frac{106.0 \, \text{g}}{1 \, \text{mol Na}_2\text{CO}_3}
\]
2. **Calculation:**
\[
\text{mass of Na}_2\text{CO}_3 = 2.5 \, \text{kg} \times 1000 \, \text{g/kg} \times \frac{1 \, \text{mol HNO}_3}{63.0 \, \text{g}} \times \frac{1 \, \text{mol Na}_2\text{CO}_3}{2 \, \text{mol HNO}_3} \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feb62ec4b-b625-4332-acd2-c86081f32c3c%2F6611826d-f039-4c72-b0a5-7fcc59abf995%2Fjshjwkg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 54: Neutralization of Nitric Acid Spill**
In an accident, a solution containing 2.5 kg of nitric acid (HNO₃) was spilled. Two kilograms of sodium carbonate (Na₂CO₃) was quickly spread on the area, and carbon dioxide (CO₂) was released by the reaction. Was sufficient Na₂CO₃ used to neutralize all of the acid?
**Solution:**
The chemical reaction is:
\[ \text{Na}_2\text{CO}_3 + 2\text{HNO}_3 \rightarrow 2\text{Na}^+ + 2\text{NO}_3^- + 2\text{CO}_2 + \text{H}_2\text{O} \]
To calculate the mass of Na₂CO₃ required for the complete reaction with nitric acid, follow these steps:
1. **Mass of Acid → Moles of Acid → Moles of Na₂CO₃ → Mass of Na₂CO₃**
- Calculate the number of moles of HNO₃:
\[
\frac{2.5 \, \text{kg} \times 1000 \, \text{g/kg}}{63.0 \, \text{g/mol}} = \text{moles of } \text{HNO}_3
\]
- Use the stoichiometry of the balanced reaction:
\[
\text{1 mol Na}_2\text{CO}_3 \text{ is required per 2 mol HNO}_3
\]
- Convert moles of Na₂CO₃ to mass:
\[
\times \frac{106.0 \, \text{g}}{1 \, \text{mol Na}_2\text{CO}_3}
\]
2. **Calculation:**
\[
\text{mass of Na}_2\text{CO}_3 = 2.5 \, \text{kg} \times 1000 \, \text{g/kg} \times \frac{1 \, \text{mol HNO}_3}{63.0 \, \text{g}} \times \frac{1 \, \text{mol Na}_2\text{CO}_3}{2 \, \text{mol HNO}_3} \
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