C. Johannesson

Ch. 10 & 11 - Ch. 10 & 11 - Gases Gases

Gas Stoichiometry at Non-STP Conditions

Gas Stoichiometry at Non-STP Conditions

C. Johannesson

A. Gas StoichiometryA. Gas StoichiometryA. Gas StoichiometryA. Gas Stoichiometry Moles Moles Liters of a Gas: Liters of a Gas:

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

Non-Non-STPSTP• Given liters of gas?

start with ideal gas law• Looking for liters of gas?

start with stoichiometry conv.

C. Johannesson

1 molCaCO3

100.09g CaCO3

B. Gas Stoichiometry B. Gas Stoichiometry ProblemProblemB. Gas Stoichiometry B. Gas Stoichiometry ProblemProblem

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

5.25 gCaCO3 = 1.26 mol CO2

CaCO3 CaO + CO2

1 molCO2

1 molCaCO3

5.25 g ? Lnon-STPLooking for liters: Start with stoich

and calculate moles of CO2.

Plug this into the Ideal Gas Law to find liters.

C. Johannesson

WORK:

PV = nRT

V =

B. Gas Stoichiometry B. Gas Stoichiometry ProblemProblemB. Gas Stoichiometry B. Gas Stoichiometry ProblemProblem

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

GIVEN:

P = 103 kPa = ? atmV = ?

n = 1.26 molT = 25°C = 298 KR = .0821 L atm/mol K

C. Johannesson

WORK:

PV = nRT

( atm) (15.0 L)= n (.0821) (294K)

n = 0.597 mol O2

B. Gas Stoichiometry B. Gas Stoichiometry ProblemProblemB. Gas Stoichiometry B. Gas Stoichiometry ProblemProblem

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

GIVEN:

P = 97.3 kPa V = 15.0 L

n = ?T = 21°C = 294 KR = .0821

4 Al + 3 O2 2 Al2O3 15.0 L

non-STP ? gGiven liters: Start with

Ideal Gas Law and calculate moles of O2.

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C. Johannesson

2 mol Al2O3

3 mol O2

B. Gas Stoichiometry B. Gas Stoichiometry ProblemProblemB. Gas Stoichiometry B. Gas Stoichiometry ProblemProblem

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

0.597mol O2 = 40.6 g Al2O3

4 Al + 3 O2 2 Al2O3

101.96 g Al2O3

1 molAl2O3

15.0Lnon-STP

? gUse stoich to convert moles of O2 to grams Al2O3.