Download presentation

Presentation is loading. Please wait.

Published byEmily Robertson Modified over 3 years ago

1
Ch Gases II. Ideal Gas Law Ideal Gas Law and Gas Stoichiometry

2
Part 1 Ideal Gas Law

3
**Standard Temperature & Pressure**

A. Avogadro’s Principle Molar Volume at STP 1 mol of a gas=22.4 L at STP Standard Temperature & Pressure 0°C and 1 atm

4
**A. Avogadro’s Principle**

Equal volumes of gases contain equal numbers of moles at constant temp & pressure true for any gas n = number of moles V n

5
**UNIVERSAL GAS CONSTANT**

B. Ideal Gas Law Merge the Combined Gas Law with Avogadro’s Principle: V n PV nT PV T = k = R UNIVERSAL GAS CONSTANT R= Latm/molK R=8.315 dm3kPa/molK

6
**UNIVERSAL GAS CONSTANT**

B. Ideal Gas Law PV=nRT UNIVERSAL GAS CONSTANT R= Latm/molK R=8.315 dm3kPa/molK

7
**C. Ideal Gas Law Problems**

Calculate the pressure in atmospheres of mol of He at 16°C & occupying L. GIVEN: P = ? atm n = mol T = 16°C = 289 K V = 3.25 L R = Latm/molK WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K P = 3.01 atm

8
**C. Ideal Gas Law Problems**

Find the volume of 85 g of O2 at 25°C and kPa. GIVEN: V = ? n = 85 g T = 25°C = 298 K P = kPa R = dm3kPa/molK WORK: 85 g 1 mol O2 = 2.7 mol 32.00 g O2 = 2.7 mol PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K V = 64 dm3

9
**D. Applications of Ideal Gas Law**

Can be used to calculate the molar mass of a gas from the density Substitute this into ideal gas law And m/V = d in g/L, so

10
**D. Applications of Ideal Gas Law**

The density of a gas was measured at atm and 27°C and found to be 1.95 g/L. Calculate the molar mass of the gas. GIVEN: P = 1.50 atm T = 27°C = 300. K d = 1.95 g/L R = Latm/molK MM = ? WORK: MM = dRT/P MM=(1.95)( )(300.)/1.50 g/L Latm/molK K atm MM = 32.0 g/mol

11
**D. Applications of Ideal Gas Law**

Calculate the density of carbon dioxide gas at 25°C and 750. torr. GIVEN: d = ? g/L CO2 T = 25°C = 298 K P = 750. torr R = Latm/molK MM = g/mol WORK: 750 torr 1 atm = .987 atm torr MM = dRT/P →d = MM P/RT d=(44.01 g/mol)(.987 atm) ( Latm/molK )(298K) d = 1.78 g/L CO2 = .987 atm

12
Part 2 Gas Stoichiometry

13
*** Stoichiometry Steps Review ***

1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. Mole ratio - moles moles Molar mass - moles grams Molarity - moles liters soln Molar volume - moles liters gas Mole ratio - moles moles Core step in all stoichiometry problems!! 4. Check answer.

14
**Standard Temperature & Pressure**

A. Molar Volume at STP 1 mol of a gas=22.4 L at STP Standard Temperature & Pressure 0°C and 1 atm

15
**A. Molar Volume at STP LITERS OF GAS AT STP Molar Volume MASS IN GRAMS**

(22.4 L/mol) MASS IN GRAMS MOLES NUMBER OF PARTICLES Molar Mass (g/mol) 6.02 1023 particles/mol

16
**B. Gas Stoichiometry Liters of one Gas Liters of another Gas:**

Avogadro’s Principle Coefficients give mole ratios and volume ratios Moles (or grams) of A Liters of B: STP – use 22.4 L/mol Non-STP – use ideal gas law & stoich Non-STP Given liters of gas? start with ideal gas law Looking for liters of gas? start with stoichiometry conv

17
**C. Gas Stoichiometry - Volume**

What volume of oxygen is needed for the complete combustion of 4.00 L of propane (C3H8)? C3H O2 CO2+ H2O 4.00 L ?L 5 L O2 1 L C3H8 4.00 L C3H8 = 20.0 L O2

18
**C. Gas Stoichiometry Problem – STP**

How many grams of KClO3 are req’d to produce 9.00 L of O2 at STP? 2KClO3 2KCl + 3O2 ? g 9.00 L 9.00 L O2 1 mol O2 22.4 L 2 mol KClO3 3 mol O2 122.55 g KClO3 1 mol KClO3 = 32.8 g KClO3

19
**D. Gas Stoichiometry Problem – Non-STP**

What volume of CO2 forms from g of CaCO3 at 103 kPa & 25ºC? CaCO3 CaO CO2 5.25 g ? L non-STP Looking for liters: Start with stoich and calculate moles of CO2. P = 103 kPa V = ? n = ? R = dm3kPa/molK T = 298K 5.25 g CaCO3 1 mol CaCO3 100.09g 1 mol CO2 CaCO3 NEXT = mol CO2 Plug this into the Ideal Gas Law for n to find liters

20
**D. Gas Stoichiometry Problem – Non-STP**

What volume of CO2 forms from g of CaCO3 at 103 kPa & 25ºC? GIVEN: P = 103 kPa V = ? n = mol T = 25°C = 298 K R = dm3kPa/molK WORK: PV = nRT (103 kPa)V =(0.0525mol)(8.315dm3kPa/molK) (298K) V = 1.26 dm3 CO2

21
**B. Gas Stoichiometry Problem**

How many grams of Al2O3 are formed from L of O2 at 97.3 kPa & 21°C? 4 Al O2 2 Al2O3 15.0 L non-STP ? g GIVEN: P = 97.3 kPa V = 15.0 L n = ? T = 21°C = 294 K R = dm3kPa/molK WORK: PV = nRT (97.3 kPa) (15.0 L) = n (8.315dm3kPa/molK) (294K) n = mol O2 Given liters: Start with Ideal Gas Law and calculate moles of O2. NEXT

22
**B. Gas Stoichiometry Problem**

How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C? 4 Al O2 2 Al2O3 15.0L non-STP ? g Use stoich to convert moles of O2 to grams Al2O3 0.597 mol O2 2 mol Al2O3 3 mol O2 g Al2O3 1 mol Al2O3 = 40.6 g Al2O3

Similar presentations

OK

C. Johannesson Ch. 10 & 11 - Gases Gas Stoichiometry at Non- STP Conditions.

C. Johannesson Ch. 10 & 11 - Gases Gas Stoichiometry at Non- STP Conditions.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on say no to drugs Ppt on voice operated intelligent fire extinguisher Ppt on ac to dc rectifier circuits Ppt on theme in literature Ppt on manufacturing of turbo generators manufacturers Ppt on coalition government big Download ppt on strings in c language Free ppt on brain-machine interface technology texting by thinking Ppt on gps based vehicle root tracking system Ppt on refraction of light through glass slab