Question

Ammonia gas reacts with oxygen gas according to the following equation: 4 NH3 + 5 O2 -----> 4 NO + 6 H2O 21. How many liters of oxygen gas are needed to react with 25 g of ammonia at STP? 22. Starting with the 25g of ammonia, what volume of nitrogen oxide was collected at the current lab temperature of 25° C and 0.943 atm? 23. Under the same conditions, how many grams of water were produced?

          Ammonia gas reacts with oxygen gas according to the following equation: 4 NH3 + 5 O2 -----> 4 NO + 6 H2O

21. How many liters of oxygen gas are needed to react with 25 g of ammonia at STP?

22. Starting with the 25g of ammonia, what volume of nitrogen oxide was collected at the current lab temperature of 25° C and 0.943 atm?

23. Under the same conditions, how many grams of water were produced?

Show more…

Ammonia gas reacts with oxygen gas according to the following equation: 4 NH3 + 5 O2 —–> 4 NO + 6 H2O

21. How many liters of oxygen gas are needed to react with 25 g of ammonia at STP?

22. Starting with the 25g of ammonia, what volume of nitrogen oxide was collected at the current lab temperature of 25° C and 0.943 atm?

23. Under the same conditions, how many grams of water were produced?

Added by Stacie R.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Madhur L

08/18/2023

Step 1

The molar mass of NH3 is 14.01 (N) + 3 * 1.01 (H) = 17.04 g/mol. So, the number of moles of NH3 is 25 g / 17.04 g/mol = 1.467 moles. Now, let's use the balanced chemical equation to find the moles of oxygen (O2) needed: 4 NH3 + 5 O2 -> 4 NO + 6 H2O From the Show more…

Show all steps

lock

Thanks for your feedback!

Ammonia gas reacts with oxygen gas according to the following equation: 4 NH3 + 5 O2 → 4 NO + 6 H2O. 21. How many liters of oxygen gas are needed to react with 25 g of ammonia at STP? 22. Starting with the 25 g of ammonia, what volume of nitrogen oxide was collected at the current lab temperature of 25 °C and 0.943 atm? 23. Under the same conditions, how many grams of water were produced?

Play audio

Feedback

Powered by NumerAI

Madhur L and 85 other subject Chemistry 101 educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Recommended Videos

Recommended Textbooks

Transcript

00:01 Hello student in this question we have to solve multiple parts.

00:03 In first part we have to calculate how many liters of oxygen gas are needed.

00:09 So volume of oxygen gas needed for this.

00:17 First of all we will calculate moles of nh3.

00:24 So this will be mass upon molar mass.

00:30 We will calculate the values that is 25 divided by 17 .03.

00:36 This is approximately 1 .468 mole.

00:41 Now from this we can see molar ratio is 4 is to 5.

00:53 Therefore moles of o2 will be n of nh3 multiplied by 5 upon 4 that is 1 .468 multiplied by 5 by 4.

01:08 This is equal to 1 .835 mole.

01:19 Now we will use pv is equal to nrt to calculate the volume.

01:24 We will rearrange to calculate for v.

01:26 V is equal to nrt upon p.

01:29 So we will substitute the values that is 1 .835 multiplied by 0 .08206.

01:44 This whole is to be multiplied by 273 .15.

01:49 Now this will be divided by 1 atm.

01:54 So on calculating it comes as 41 .27 liter.

02:02 So volume of o2 is equal to 41 .27 liter.

02:10 Now b part what volume of nitrogen oxide was collected...

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace

Continue solving this problem

Create a free account to:

View full step-by-step solution
Ask follow-up questions with Ace AI
Save progress and study later