Class opener tuesday march 18
This presentation is the property of its rightful owner.
Sponsored Links
1 / 7

Class Opener : Tuesday, March 18 PowerPoint PPT Presentation


  • 52 Views
  • Uploaded on
  • Presentation posted in: General

Class Opener : Tuesday, March 18. 1. The reaction shown represents the oxidation of ammonia (NH 3 ). 4 NH 3 + 5 O 2  4NO + 6 H 2 O How many moles of nitrogen monoxide (NO) would be formed from 34 grams of ammonia?

Download Presentation

Class Opener : Tuesday, March 18

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Class opener tuesday march 18

Class Opener:Tuesday, March 18

1.The reaction shown represents the oxidation of ammonia (NH3).

4 NH3 + 5 O2  4NO + 6 H2O

How many moles of nitrogen monoxide (NO) would be formed from 34 grams of ammonia?

2.If this reaction took place at STP, what would be the volume (in liters) of the NO formed?


Comparing the volumes of reacting gases

Comparing the Volumes of Reacting Gases

  • In the early 1800s, Gay-Lussac observed that when gases reacted, they did so in simple and definite volume proportions.

  • Example:

    hydrogen gas + oxygen gas → water vapor 2 L (2 volumes) 1 L (1 volume) 2 L (2 volumes)

  • Gay-Lussac’s law of combining volumes of gases states that at constant temperature and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers.


Class opener tuesday march 18

  • The coefficients in chemical equations of gas reactions reflect not only molar ratios, but also volume ratios.

    • Example:

      2CO(g) + O2(g) → 2CO2(g)

      2 mole 1 mole 2 mol

      2 volumes 1 volume 2 volumes


Sample problem

Sample Problem

The complete combustion of propane, C3H8, occurs according to the following balanced equation.

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

How many liters of oxygen (O2) will be required for the complete combustion of 0.350 L of propane?


Gas stoichiometry

Gas Stoichiometry

The industrial production of ammonia (NH3) proceeds according to the following equation:

N2(g) + 3H2(g) → 2 NH3(g)

If 20.0 mol of nitrogen (N2) is available, how many liters of NH3 can be produced at STP?


Eoc sample question

EOC Sample Question

The equation represents the breakdown of potassium chlorate (KClO3).

2 KClO3 2 KCl + 3 O2

What volume of oxygen gas (O2) does 20.0 grams of potassium chlorate (KClO3) produce at STP based on the equation shown?

A. 5.48 L

B. 7.80 L

C. 67.3 L

D. 72.9 L


Wrapping it up

Wrapping It Up

Reflect on what you have learned concerning the today’s topic.

Respond to the following…

What is the most important thing you learned today?

What is one question you would still like answered?

What is a way what you have learned today connects with what you knew before?


  • Login