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Get Full Access to Introductory Chemistry - 5 Edition - Chapter 8 - Problem 17p
Get Full Access to Introductory Chemistry - 5 Edition - Chapter 8 - Problem 17p

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# For the reaction shown, calculate how many moles of NO2 form when each amount of

ISBN: 9780321910295 34

## Solution for problem 17P Chapter 8

Introductory Chemistry | 5th Edition

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Problem 17P

For the reaction shown, calculate how many moles of $$\mathrm{NO}_{2}$$ form when each amount of reactant completely reacts.

$$2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \rightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$$

(a) 1.3 mol $$\mathrm{N}_{2} \mathrm{O}_{5}$$

(b) 5.8 mol $$\mathrm{N}_{2} \mathrm{O}_{5}$$

(c) $$4.45 \times 10^{3} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{5}$$

(d) $$1.006 \times 10^{-3} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{5}$$

Equation Transcription:

Text Transcription:

NO_2

2 N_2 O_5 (g) right arrow 4 NO_2 (g) + O_2 (g)

N_2 O_5

4.45 times 10^3 mol N_2 O_5

1.006 times 10^-3 mol N_2 O_5

Step-by-Step Solution:

Solution 17P

Here, we are going to calculate the number of moles of NO2 formed when the given amounts of N2O5 completely reacts.

Step 1:

The given chemical reaction is:

2N2O5(g) → 4NO2(g) + O2(g)

From the above equation, it is clear that, 2 moles of N2O5 reacts completely to give 4 moles of NO2 as product.

Therefore, 1 mole of N2O5 will give (4/2 = 2) mol of NO2 on complete reaction.

Step 2 of 2

##### ISBN: 9780321910295

This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. The full step-by-step solution to problem: 17P from chapter: 8 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. This full solution covers the following key subjects: mol, Moles, completely, form, amount. This expansive textbook survival guide covers 19 chapters, and 2046 solutions. Since the solution to 17P from 8 chapter was answered, more than 3616 students have viewed the full step-by-step answer. The answer to “?For the reaction shown, calculate how many moles of $$\mathrm{NO}_{2}$$ form when each amount of reactant completely reacts.$$2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \rightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$$(a) 1.3 mol $$\mathrm{N}_{2} \mathrm{O}_{5}$$(b) 5.8 mol $$\mathrm{N}_{2} \mathrm{O}_{5}$$(c) $$4.45 \times 10^{3} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{5}$$(d) $$1.006 \times 10^{-3} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{5}$$Equation Transcription:Text Transcription:NO_22 N_2 O_5 (g) right arrow 4 NO_2 (g) + O_2 (g)N_2 O_54.45 times 10^3 mol N_2 O_51.006 times 10^-3 mol N_2 O_5” is broken down into a number of easy to follow steps, and 67 words. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295.

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