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Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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![**Part F**
Calculate the grams of NO that would be produced.
Express your answer with the appropriate units.
\[ m(\text{NO}) = \, \text{Value} \, \text{Units} \]
[Submit] [Request Answer]
---
**Part G**
Determine the limiting reactant, given 31.0 g of each reactant, in the following reaction:
\[ \text{C}_2\text{H}_6\text{O}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(g) \]
---
**Explanation:**
The given equation calculates the moles of \(\text{HNO}_3\) formed:
- \[
31.0 \, \text{g} \, \text{H}_2\text{O} \times \frac{1 \, \text{mol} \, \text{H}_2\text{O}}{18.015 \, \text{g}} \times \frac{2 \, \text{mol} \, \text{HNO}_3}{1 \, \text{mol} \, \text{H}_2\text{O}} = 3.44 \, \text{mol} \, \text{HNO}_3
\]
Since the number of moles produced by NO\(_2\) is less, it is identified as the limiting reagent.](https://content.bartleby.com/qna-images/question/ec34e07b-01ec-4806-810c-ebd28802db2a/b5d8f608-c6fc-4f4f-a34c-6b205d3f2161/o4iblpp_thumbnail.jpeg)
Transcribed Image Text:**Part F**
Calculate the grams of NO that would be produced.
Express your answer with the appropriate units.
\[ m(\text{NO}) = \, \text{Value} \, \text{Units} \]
[Submit] [Request Answer]
---
**Part G**
Determine the limiting reactant, given 31.0 g of each reactant, in the following reaction:
\[ \text{C}_2\text{H}_6\text{O}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(g) \]
---
**Explanation:**
The given equation calculates the moles of \(\text{HNO}_3\) formed:
- \[
31.0 \, \text{g} \, \text{H}_2\text{O} \times \frac{1 \, \text{mol} \, \text{H}_2\text{O}}{18.015 \, \text{g}} \times \frac{2 \, \text{mol} \, \text{HNO}_3}{1 \, \text{mol} \, \text{H}_2\text{O}} = 3.44 \, \text{mol} \, \text{HNO}_3
\]
Since the number of moles produced by NO\(_2\) is less, it is identified as the limiting reagent.
![Determine the limiting reactant, given 31.0 g of each reactant, in the following reaction:
\[3NO_2(g) + H_2O(l) \rightarrow 2HNO_3(aq) + NO(g)\]
Options for the limiting reactant:
- NO
- HNO₃
- NO₂
- H₂O
---
Correct Answer Explanation:
To find the limiting reactant, use the 31.0 g of each reactant and calculate the amount of product formed by each reactant. The limiting reactant is the one that forms the least amount of product, as it will run out first. Either product can be used for this comparison. In this example, \(HNO_3\) is used.
1. **Calculation for \(NO_2\):**
\[
\text{moles of } HNO_3 \text{ formed} = 31.0 \, \cancel{g \, NO_2} \times \frac{1 \, \cancel{mol \, NO_2}}{46.00 \, \cancel{g \, NO_2}} \times \frac{2 \, mol \, HNO_3}{3 \, \cancel{mol \, NO_2}} = 0.449 \, mol \, HNO_3
\]
2. **Calculation for \(H_2O\):**
\[
\text{moles of } HNO_3 \text{ formed} = 31.0 \, \cancel{g \, H_2O} \times \frac{1 \, \cancel{mol \, H_2O}}{18.015 \, \cancel{g \, H_2O}} \times \frac{2 \, mol \, HNO_3}{1 \, \cancel{mol \, H_2O}} = 3.44 \, mol \, HNO_3
\]
Since the number of moles produced by \(NO_2\) is less, \(NO_2\) must be the limiting reagent.](https://content.bartleby.com/qna-images/question/ec34e07b-01ec-4806-810c-ebd28802db2a/b5d8f608-c6fc-4f4f-a34c-6b205d3f2161/cwklrhb_thumbnail.jpeg)
Transcribed Image Text:Determine the limiting reactant, given 31.0 g of each reactant, in the following reaction:
\[3NO_2(g) + H_2O(l) \rightarrow 2HNO_3(aq) + NO(g)\]
Options for the limiting reactant:
- NO
- HNO₃
- NO₂
- H₂O
---
Correct Answer Explanation:
To find the limiting reactant, use the 31.0 g of each reactant and calculate the amount of product formed by each reactant. The limiting reactant is the one that forms the least amount of product, as it will run out first. Either product can be used for this comparison. In this example, \(HNO_3\) is used.
1. **Calculation for \(NO_2\):**
\[
\text{moles of } HNO_3 \text{ formed} = 31.0 \, \cancel{g \, NO_2} \times \frac{1 \, \cancel{mol \, NO_2}}{46.00 \, \cancel{g \, NO_2}} \times \frac{2 \, mol \, HNO_3}{3 \, \cancel{mol \, NO_2}} = 0.449 \, mol \, HNO_3
\]
2. **Calculation for \(H_2O\):**
\[
\text{moles of } HNO_3 \text{ formed} = 31.0 \, \cancel{g \, H_2O} \times \frac{1 \, \cancel{mol \, H_2O}}{18.015 \, \cancel{g \, H_2O}} \times \frac{2 \, mol \, HNO_3}{1 \, \cancel{mol \, H_2O}} = 3.44 \, mol \, HNO_3
\]
Since the number of moles produced by \(NO_2\) is less, \(NO_2\) must be the limiting reagent.
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