Gas Stoichiometry

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Gas Stoichiometry. Moles  Liters of a Gas : STP - use 22.4 L/mol Non-STP - use ideal gas law Non- STP Given liters of gas? start with ideal gas law Looking for liters of gas? start with stoichiometry conversion. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem.

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Gas Stoichiometry

Moles  Liters of a Gas:

• STP - use 22.4 L/mol
• Non-STP - use ideal gas law

Non-STP

• Given liters of gas?
• Looking for liters of gas?

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Gas Stoichiometry Problem

What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC?

CaCO3 CaO + CO2

5.25 g

? Lnon-STP

5.25 g

CaCO3

1 mol

CaCO3

100.09g

CaCO3

1 mol

CO2

1 mol

CaCO3

= 1.26 mol CO2

Plug this into the Ideal Gas Law to find liters.

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Gas Stoichiometry Problem
• What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC?

GIVEN:

P = 103 kPa

V = ?

n = 1.26 mol

T = 25°C = 298 K

R = 8.315dm3kPa/molK

WORK:

PV = nRT

(103 kPa)V=(1mol)(8.315dm3kPa/molK)(298K)

V = 1.26 dm3 CO2

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Gas Stoichiometry Problem

How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?

4 Al + 3 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 = 8.315dm3kPa/molK

WORK:

PV = nRT

(97.3 kPa) (15.0 L)= n (8.315dm3kPa/molK) (294K)

n = 0.597 mol O2

Given liters: Start with Ideal Gas Law and calculate moles of O2.

NEXT 

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Gas Stoichiometry Problem

How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?

4 Al + 3 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

101.96 g Al2O3

1 mol

Al2O3

= 40.6 g Al2O3

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

n R T

V =

P

Gas Stoichiometry

Find vol. hydrogen gas made when 38.2 g zinc react w/excess hydrochloric acid.

Pres. = 107.3 kPa; temp.= 88oC.

Zn (s) + 2 HCl (aq) ZnCl2(aq) + H2(g)

38.2 g

excess

X L

P = 107.3 kPa

(13.1 L)

T = 88oC

1 mol H2

22.4 L O2

1 mol Zn

x L H2 = 38.2 g Zn

= 13.1 L H2

1 mol Zn

1 mol H2

65.4 g Zn

Zn

H2

Combined

Gas Law

At STP, we’d use 22.4 L per 1 mol, but we aren’t at STP.

1 mol H2

1 mol Zn

x mol H2 = 38.2 g Zn

= 0.584 mol H2

1 mol Zn

65.4 g Zn

88oC + 273 = 361 K

0.584 mol (8.314 L.kPa/mol.K)(361 K)

P V = n R T

16.3 L

=

=

107.3 kPa

Gas Stoichiometry

Find vol. hydrogen gas made when 38.2 g zinc react w/excess hydrochloric acid.

Pres. = 107.3 kPa; temp.= 88oC.

Zn (s) + 2 HCl (aq) ZnCl2(aq) + H2(g)

38.2 g

excess

X L

P = 107.3 kPa

(13.1 L)

T = 88oC

1 mol H2

22.4 L O2

1 mol Zn

x L H2 = 38.2 g Zn

= 13.1 L H2

1 mol Zn

1 mol H2

65.4 g Zn

Zn

H2

Combined

Gas Law

At STP, we’d use 22.4 L per 1 mol, but we aren’t at STP.

P1 =

T1 =

V1 =

P2 =

T2 =

V2 =

101.3 kPa

(101.3 kPa) x (13.1 L) = (107.3 kPa) x (V2)

P1 x V1

T1

P2 x V2

T2

=

273 K

273 K

361 K

13.1 L

107.3 kPa

16.3 L

V2

=

88 oC

+ 273

= 361 K

X L

P V

n =

R T

What mass solid magnesium is required to react w/250 mL carbon dioxide

at 1.5 atm and 77oC to produce solid magnesium oxide and solid carbon?

2

+ CO2 (g)

2

MgO (s)

+ C (s)

Mg (s)

250 mL

0.25 L

X g Mg

0.25 L

V = 250 mL

oC + 273 = K

T = 77oC

350 K

P = 1.5 atm

151.95 kPa

(0.250 L)

151.95 kPa

1.5 atm

= 0.013 mol CO2

n =

P V = n R T

(350 K)

0.0821 L.atm / mol.K

8.314 L.kPa / mol.K

24.3 g Mg

2 mol Mg

x g Mg = 0.013 mol CO2

= 0.63 g Mg

1 mol Mg

1 mol CO2

CO2

Mg

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?

2

2

Na + Cl2 NaCl

excess

X L

5 g

1 mol NaCl

1 mol Cl2

22.4 L Cl2

x g Cl2 = 5 g NaCl

= 0.957 L Cl2

58.5 g NaCl

2 mol NaCl

1 mol Cl2

P1 x V1

T1

P2 x V2

T2

Ideal Gas

Method

=

P1 = 1 atm

T1 = 273 K

V1 = 0.957 L

P2 = 0.95 atm

T2 = 25 oC + 273 = 298 K

V2 = X L

(1 atm) x (0.957 L) (0.95 atm) x (V2)

=

273 K

298 K

V2 = 1.04 L

n R T

V =

P

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?

2

2

Na + Cl2 NaCl

excess

X L

5 g

1 mol NaCl

1 mol Cl2

x g Cl2 = 5 g NaCl

= 0.0427 mol Cl2

58.5 g NaCl

2 mol NaCl

Ideal Gas

Method

P V = n R T

P = 0.95 atm

T = 25 oC + 273 = 298 K

V = X L

R = 0.0821 L.atm / mol.K

n = 0.0427 mol

0.0427 mol (0.0821 L.atm / mol.K) (298 K)

X L =

0.95 atm

V = 1.04 L