- 119 Views
- Uploaded on
- Presentation posted in: General

Gas Stoichiometry Ideal Gas Law

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 - - - - - - - - - - - - - - - - - - - - - - - - - -

Gas Stoichiometry Ideal Gas Law

Putting it all together…

- C8H18(l) + O2(g) → CO2(g) + H20(g)
- 2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H20(g)

- Problem 11.13 c (p.332)
- Mg + HCl → H2 + MgCl2
- 60.1 g HCl

- Problem 12.85 (p.373)
- 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

- This means that the mole ratio is the same as the volume ratio

- Problem 12.85 (p.373)
- 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
- 200. L O2

- Problem 12.85 (p.373)
- 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

- Problem 12.81 (p.373)
- 2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2H2O(g)
- 2.5 tanks O2

- Problem 12.92 (p374)
- NaN3(s) + Fe2O3(s) → Na2O(s) + Fe(s) + N2(g)
- 6 NaN3(s) + Fe2O3(s) → 3 Na2O(s) + 2 Fe(s) + 9 N2(g)
- 10.6 g NaN3

- The volume of 1.0 mole of any gas at STP is 22.4 L.
- This is called the standard molar volume.

- The volume of any gas at STP can be calculated if the number of moles is known:
- V = (moles)(22.4 L)

- Problem 12.88 (p.373)
- CaC2(s) + 2 H2O(l) → Ca(OH)2(s) + C2H2(g)

- All of the variables used to describe gases can be compared using the Ideal Gas Law
- The IGL is often called an equation of state, as it is very useful when finding some property of a gas at a certain condition
- Not so great when conditions are changing
- Entropy is another condition of state

- PV = nRT
- P = pressure
- V = volume
- T = temperature
- n = number of moles of gas
- R = Gas Constant
- value depends on the desired unit

- R is a constant that is used in the IGL so that it may be used for all gases
- It has a different value, depending on the pressure unit being used in the problem

- Problem 12.88 (p.373)
- CaC2(s) + 2 H2O(l) → Ca(OH)2(s) + C2H2(g)
- 960 ml C2H2

- Problem 12.90 (p.373)
- 2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H20(g)

- Problem 12.88 (p.373)
- CaC2(s) + 2 H2O(l) → Ca(OH)2(s) + C2H2(g)
- 960 ml C2H2

- Problem 12.90 (p.373)
- 2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H20(g)
- 837 L CO2