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Gas Stoichiometry Ideal Gas LawPowerPoint Presentation

Gas Stoichiometry Ideal Gas Law

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## PowerPoint Slideshow about ' Gas Stoichiometry Ideal Gas Law' - maverick-robertson

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### Gas Stoichiometry Ideal Gas Law

Putting it all together…

Balance the following equation… Coefficients of balanced equations can be used as mole ratios in stoichiometry problems

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

Mass-Mass Stoichiometry

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

Volume Ratio = Mole Ratio

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

Avogadro’s LawOne mole of a gas will occupy 22.4 L

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

Volume Ratio = Mole Ratio

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

Volume Ratio = Mole Ratio

- 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

Gas @ STP

- 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

Standard Molar Volume

- 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)

Gas Not at STP

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

Ideal Gas Law

- 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

Ideal Gas Law

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

Gas Constant (“R”)

- 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

Gas Not at STP

- 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)

Gas Not at STP

- 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

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