4.119 If the % yield of nitrogen gas in Question 4.115 is 75.0%, what is the actual yield of nitrogen?4.120 If the % yield of Fe₂O₃ in Question 4.116 is 90.0%, what is the actual yield of Fe₂O₃? **4.115** A rocket can be powered by the reaction between dinitrogen tetroxide and hydrazine:\[ \text{N}_2\text{O}_4 (l) + 2\text{N}_2\text{H}_4 (l) \rightarrow 3\text{N}_2 (g) + 4\text{H}_2\text{O} (g) \]An engineer designed the rocket to hold 1.00 kg $\text{N}_2\text{O}_4$ and excess $\text{N}_2\text{H}_4$. How many g of $\text{N}_2$ would be produced according to the engineer’s design?**4.116** A 4.00-g sample of $\text{Fe}_3\text{O}_4$ reacts with $\text{O}_2$ to produce $\text{Fe}_2\text{O}_3$:\[ 4\text{Fe}_3\text{O}_4 (s) + \text{O}_2 (g) \rightarrow 6\text{Fe}_2\text{O}_3 (s) \]Determine the number of g of $\text{Fe}_2\text{O}_3$ produced.

Answered: 115 A rocket can be powered by the…

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

Related questions

Concept explainers

Question

**4.115** A rocket can be powered by the reaction between dinitrogen tetroxide and hydrazine:

\[ \text{N}_2\text{O}_4 (l) + 2\text{N}_2\text{H}_4 (l) \rightarrow 3\text{N}_2 (g) + 4\text{H}_2\text{O} (g) \]

An engineer designed the rocket to hold 1.00 kg $\text{N}_2\text{O}_4$ and excess $\text{N}_2\text{H}_4$. How many g of $\text{N}_2$ would be produced according to the engineer’s design?

**4.116** A 4.00-g sample of $\text{Fe}_3\text{O}_4$ reacts with $\text{O}_2$ to produce $\text{Fe}_2\text{O}_3$:

\[ 4\text{Fe}_3\text{O}_4 (s) + \text{O}_2 (g) \rightarrow 6\text{Fe}_2\text{O}_3 (s) \]

Determine the number of g of $\text{Fe}_2\text{O}_3$ produced.

Expert Solution

Step 1

4.115) Given,

Nitrogen tetraoxide reacts with hydrazine to form nitrogen gas and water

$N_{2} O_{4} (l) + 2 N_{2} H_{4} \to 3 N_{2} + 4 H_{2} O$

Mass of N₂O₄ = 1 kg= 1000 g

Molar mass of nitrogen tetraoxide (N₂O₄) is = 92.011 g/mol

Molar mass of N₂ is = 28.0134 g/mol

From the stiochiometry of the reaction one mole of nitrogen tetraoxide reacts with two moles of hydrazine to form three moles of nitrogen and four moles of water.

$1 m o l e o f N_{2} O_{4} f o r m s = 3 m o l e s o f N_{2}$

$92.011 g o f N_{2} O_{4} f o r m s = 3 \times 28.0134 g o f N_{2}$

$1 g o f N_{2} O_{4} f o r m s = \frac{3 \times 28.0134}{92.011} o f N_{2}$

$1000 g o f N_{2} O_{4} f o r m s = \frac{3 \times 28.0134}{92.011} \times 1000 g o f N_{2}$

$1000 g o f N_{2} O_{4} f o r m s = 913.4 g o f N_{2}$

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions