21st Century Literature 21st Century Literature 21st Century Literature, Summaries of Chemistry

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Opening Prayer

Starting Activity

You are given a dare to breath-in a balloon full of helium and then

speak to your friends after you have done it.

1. What do you think will happen next?

2. Why do our voices change when we breath in different gases?

Learning Targets

Apply the principles of stoichiometry to determine the amounts

(volume, number of moles, or mass) of gaseous reactants and

products.

Relate the rate of gas effusion with molar mass.

Gas Stoichiometry

Two Types of Gas Stoichiometry Problems

1. STP gases (STP - use 22.4 L/mol)

2. Non-STP gases (Non-STP - use ideal gas law)

Moles  Liters of a Gas:

◦ STP - use 22.4 L/mol

◦ Non-STP - use ideal gas law

◦ *STP = 0°C @ 1 atm

Non-STP

Given liters of gas?

start with ideal gas law

Looking for liters of gas?

start with stoichiometry conversion

Gas Stoichiometry

What to do?

1. Write a reaction.

2. Find moles (gram-mole conversions, molarity, and ideal gas law)

3. Do a stoichiometric conversion between chemicals

4. Answer the question

1. 22.4 L/mol for STP gases
2. PV = nRT for non-STP gases

Gas Stoichiometry

Constants Pressure (P) Atmosphere (atm) Kilopascal (kPa) 1 atm = 101.3 kPa 1 atm = 760 mm Hg 1 atm = 760 torr Temperature (T) Kelvin ONLY To convert to Kelvin, simply add 273. Volume (V) Liter (L) Milliliter (mL) 1000 mL = 1L Moles (n) =6.0221 x10^ = g/mols = 22.4L/mol

Gas Stoichiometry

Constants

1 mol

CaCO

3

100.09g

CaCO

3

Gas Stoichiometry Problem

What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC?

5.25 g

CaCO

3

= 1.26 mol CO

2

CaCO

3

 CaO + CO

2

1 mol

CO

2

1 mol

CaCO

3

5.25 g? L

Looking for liters: Start with stoich and non-STP

calculate moles of CO 2 . Plug this into the Ideal Gas Law to find liters.

WORK:

PV = nRT

(1.0168 atm)V

=(1.26 mol)(0.08206 Latm/molK)(298K)

V = 1.17 L of CO

2

Gas Stoichiometry Problem

What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC?

GIVEN:

P = 103 kPa to atm =

1.0168 atm

V =?

n = 1.26 mol

T = 25°C = 298 K

R = 0.08206 Latm/molK

WORK:

PV = nRT

(0.961 atm) (15.0 L)

= n (0.08206 Latm/molK) (294K)

n = 0.597 mol O

2

Gas Stoichiometry Problem

How many grams of Al

2

O

3

are formed from 15.0 L of O

2

at

97.3 kPa & 21°C?

GIVEN:

P = 97.3 kPa

V = 15.0 L

n =? mol

T = 21°C = 294.15 K

R = 0.08206 Latm/molK

4 Al + 3 O

2

 2 Al

2

O

3 15.0 L

non-STP? g

Given liters: Start with Ideal Gas Law and calculate moles of O 2.

2 mol Al

2

O

3

3 mol O

2

Gas Stoichiometry Problem

How many grams of Al 2 O 3 are formed from 15. L of O 2 at 97.3 kPa & 21°C?

0.597 mol O

2

= 40.614 g Al

2

O

3

4 Al + 3 O

2

 2 Al

2

O

3

101.96 g

Al

2

O

3

1 mol

Al

2

O

3

15.0L

non-STP

Use stoic to convert moles^? g

of O 2 to grams Al 2 O 3.

Gas Stoichiometry Problem

What volume of hydrogen gas will form at STP when 30.0 g of

sodium react with an excess of hydrochloric acid, HCl?

__Na + __HCl ____NaCl + ___ H

Given: (@ STP)

30g Na

HCl = Excess

H2 =? L

Gas Stoichiometry Problem

What volume of hydrogen gas will form when 30.0 g of sodium react with an excess of hydrochloric acid, HCl @ 3 atm and 5°C? __Na + __HCl ____NaCl + ___ H Given: (@ STP) 30g Na HCl = Excess H2 =? L

Gas Stoichiometry

How many liters of chlorine gas are needed to react with excess sodium metal to yield 5.0 g of sodium chloride when T = 25oC and P = 0.95 atm? Na + Cl 2 (^2 2) NaCl excess X L 5 g x g Cl 2 = 5 g NaCl 1 mol NaCl 58.5 g NaCl 2 mol NaCl 1 mol Cl 2 22.4 L Cl 2 1 mol Cl 2 = 0.957 L Cl 2 P 1 = 1 atm T 1 = 273 K V 1 = 0.957 L P 2 = 0.95 atm T 2 = 25 o C + 273 = 298 K V 2 = X L

=

P

2 x V 2 T 2

P

1 x V 1 T 1 (1 atm) x (0.957 L) (0.95 atm) x (V 2 ) 273 K 298 K V 2

= 1.04 L

= Ideal Gas Method

Gas Stoichiometry

How many liters of chlorine gas are needed to react with excess sodium metal to yield 5.0 g of sodium chloride when T = 25oC and P = 0.95 atm? Na + Cl 2 (^2 2) NaCl excess X L 5 g x g Cl 2 = 5 g NaCl 1 mol NaCl 58.5 g NaCl 2 mol NaCl 1 mol Cl 2 = 0.0427 mol Cl 2 P = 0.95 atm T = 25 oC + 273.15 = 298 K V = X L R = 0.0821 L.atm / mol.K n = 0.0427 mol P V = n R T 0.0427 mol (0.0821 L . atm / mol . K) (298.15 K) V = 1.04 L

V =

n R T P 0.95 atm Ideal Gas Method X L =

Graham’s Law of Effusion

Based from the Kinetic molecular theory, the average kinetic energy is directly proportional to the absolute temperature.

KE(A) = KE(B)

kT = kT

½ a(mv) = ½ b(mv)

½ a(nmv) = ½ b(nmv)

Va/Vb

Where v = velocity of molecules at a given temperature.

Graham’s Law of Effusion

Based from the Kinetic molecular theory, the average kinetic energy is directly proportional to the absolute temperature.

KE(A) = KE(B)

kT = kT

½ a(mv) = ½ b(mv)

r^1/r^2 = √m2/m

Where r =rate of effusion and m are the molar masses

Graham’s Law of Effusion

What is the rate of effusion between Helium and Argon Solution: Get the Molar mass of each He = 4 Ar = 39. =r1/r2 = m2/m1 (substitute) = r1/r2 = (39.95)/(4) (Divide and square root) = r1 = 3.

Scaffold 4

Subject Reflection: write a reflective journal on your notebook based on the given question and upload a copy of your work o your Genyo Account.

1. What was your most favorite topic to learn in our subject?
2. What was the most difficult task that you are able to accomplish?
3. What is the most significant lesson that you have been taught in our subject?

Offline Activity

Solve for the following: (5 pts. Each – Show your solution and upload your answers in a word document.)

1. What volume of hydrogen gas will form at STP when 50.0 g of sodium react with an excess of hydrochloric acid, HCl?
2. What mass of potassium chlorate will be needed to make 500. mL of oxygen at 25.0 o C and 800. torr? KClO3 KCl + O2
3. What volume (L) of oxygen will form when 40.0g of calcium chlorate decompose at 352 o C and 901 torr?
4. What is the rate of effusion of two gases, mainly Hydrogen and Chlorine?