Homework Help is Here – Start Your Trial Now!
Science
Chemistry
Calculate the grams of NO that would be produced. Express your answer with the appropriate units.

Calculate the grams of NO that would be produced. Express your answer with the appropriate units.

Chemistry
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...
See similar textbooks
icon
Related questions
Question
100%
Need assistance on the formula of how to answer this
**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.
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.
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.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY
SEE MORE TEXTBOOKS