Ideal Gas Equation

PV=nRT

Practice Problems

2 AlCl3 2 Al + 3 Cl2 ↑ If 13 g of aluminum chloride react, what volume of chlorine

gas will be collected at 25°C and 1.15 atm?

Plan of action: 3 3 2 2PV=nRTmolar molemass ratio

g AlCl mol AlCl mol Cl Vol Cl

313g AlCl 3

3

1mol AlCl133.3g AlCl

2

3

3 mol Cl2 mol AlCl

2= 0.146 mol Cl

nRTV = P

2L atmmol K

0.146 mol Cl 0.0821 25+273K=

1.15 atm

2= 3.1 L Cl

2 Na + H2SO4 Na2SO4 + H2 ↑ If 3.5 g of sodium react, what pressure of hydrogen gas will

result in a 250 mL container at 35°C?

Plan of action: 2 2PV=nRTmolar molemass ratio

g Na mol Na mol H Pressure H

3.5 g Na 1 mol Na22.9898g Na

21 mol H2 mol Na

2= 0.0761 mol H

nRTP = V

2L atmmol K

0.0761 mol H 0.0821 35+273K=

1 L250 mL1000 mL

2= 7.7 atm H

58.097g KF1 mol KF

F2 + 2 KBr 2 KF + Br2 If 1.0 L of fluorine gas measured at 22°C at 0.95 atm reacts, how many grams of potassium fluoride will be produced?

Plan of action: 2PV=nRT mole molarmassratio

Use P,V,T mol F mol KF g KF

PVn = RT

L atmmol K

0.95 atm 1.0 L=

0.0821 22 + 273K

2= 0.0392 mol F

20.0392 mol F2

2 mol KF1 mol F

= 4.6 g KF

2 H2 + O2 2 H2O If 3.5 L of oxygen gas measured at 42°C at 1.25 atm reacts,

how many grams of water will be produced?

2

2

18.02 g H O1 mol H O

Plan of action: 2PV=nRT 2 2mole molarmassratio

Use P,V,T mol O mol H O g H O

PVn = RT

L atmmol K

1.25 atm 3.5 L=

0.0821 42 + 273K

2= 0.169 mol O

20.169 mol O 2

2

2 mol H O1 mol O

2= 6.1 g H O

2 KNO3 2 KNO2 + O2 ↑ When 36 g of potassium nitrate decompose, what volume of

oxygen gas is collected at 39°C and 2.3 atm pressure?

Plan of action: 3 3 2 2PV=nRTmolar molemass ratio

g KNO mol KNO mol O Vol O

336 g KNO 3

3

1mol KNO101.1g KNO

2

3

1 mol O2 mol KNO

2= 0.178 mol O

nRTV = P

2L atmmol K

0.178 mol Cl 0.0821 39+273K=

2.3 atm

2= 2.0 L O

2 Ca + O2 2 CaO In order to react 58 g of calcium, what pressure of oxygen gas

at 42°C in a 6.0 L container is needed?

Plan of action: 2 2PV=nRTmolar molemass ratio

g Ca mol Ca mol O Pressure O

58 g Na 1 mol Ca40.078 g Ca

21 mol O2 mol Ca

2= 0.7236 mol O

nRTP = V

2L atmmol K

0.7236 mol O 0.0821 42+273K=

6.0 L

2= 3.1 atm O

3 Ca + 2 H3PO4 Ca3(PO4)2 + 3 H2 ↑ If 9.5 g of calcium phosphate are produced, what volume of

hydrogen gas at 77°C and 0.88 atm pressure is also produced?

Plan of action: 2 23 4 3 4 2 2PV=nRTmolar molemass ratio

Ca (PO ) Ca (PO )g mol mol H Vol H

23 4Ca (PO )9.5 g 3 4 2

3 4 2

1mol Ca (PO )310.2 g Ca (PO )

2

3 4 2

3 mol H1 mol Ca (PO )

2= 0.0919 mol H

nRTV = P

2L atmmol K

0.0919 mol H 0.0821 77+273K=

0.88 atm

2= 3.0 L H

2 H2 + O2 2 H2O When 4.7 L of hydrogen gas at 3.2 atm and 13°C are reacted with 3.6 L of oxygen gas at 2.9 atm and 12°C, what mass of

water is produced?

Let’s try some excess-limiting:

Plan of action: Use P,V,T to find n

of H2 & O2

Use mole ratio

to find Limiting Reactant

Use Limiting to find mass

of H2O

2

2

H

O

PV= RT

PV= RT

n

n L atmmol K

L atmmol K

3.2 atm 4.7 L

0.0821 13+273 K

2.9 atm 3.6 L

0.0821 12+273 K

2= 0.641 mol H

2= 0.446 mol O

2

2

1 mol O2 mol H

2

2

2 mol H1 mol O

2= 0.321 mol O

2= 0.892 mol H

HAVE NEEDLimiting

20.641 mol H 2

2

2 mol H O2 mol H

2

2

18.02 g H O1 mol H O

2= 12 g H O

N2 + 3 H2 2 NH3 ↑ When 2.2 L of nitrogen gas at 6.8 atm and 358°C are reacted with 9.6 L of hydrogen gas at 7.4 atm and 419°C, what mass

of ammonia is produced?

Here is more excess-limiting:

Plan of action: Use P,V,T to find n

of N2 & H2

Use mole ratio

to find Limiting Reactant

Use Limiting to find mass

of NH3

2

2

N

H

PV= RT

PV= RT

n

n L atmmol K

L atmmol K

6.8 atm 2.2 L

0.0821 358+273 K

7.4 atm 9.6 L

0.0821 419+273 K

2= 0.289 mol N

2= 1.25 mol H

2

2

3 mol H1 mol N

2

2

1 mol N3 mol H

2= 0.866 mol H

2= 0.417 mol N

HAVE NEEDLimiting

20.289 mol N 3

2

2 mol NH1 mol N

3

3

17.03 g NH1 mol NH

3= 9.8 g NH

C6H12O6 + 6O2 6CO2 + 6H2O

When 8.0 L of oxygen gas at 0.98 atm and 25°C are reacted with sugar as shown, what volume of CO2 is produced at 37°C

and 0.98 atm pressure?

And now, volume-volume:

Plan of action: 2 2 2 2PV=nRT PV=nRTmoleratio

Vol O mol O mol CO Vol CO

2OPV = RT

n L atmmol K

0.98 atm 8.0 L=

0.0821 25+273K

2= 0.296 mol O 2

2

6 mol CO6 mol O

2= 0.296 mol CO

nRTV = P

2L atmmol K

0.296 mol H 0.0821 37+273K=

0.98 atm

2= 7.7 L CO

C3H8 + 5O2 3CO2 + 4H2O When 14 L of propane at 9.0 atm and 25°C are

combusted with oxygen as shown, what volume of CO2 is produced at 40°C and 0.92 atm pressure?

Plan of action: 3 8 3 8 2 2PV=nRT PV=nRTmoleratio

Vol C H mol C H mol CO Vol CO

3 8C HPV= RT

n L atmmol K

9.0 atm 14 L=

0.0821 25+273K

3 8C H= 5.15 mol 2

3 8

3 mol CO1 mol C H

2= 15.5 mol CO

nRTV = P

2L atmmol K

15.5 mol CO 0.0821 40+273K=

0.92 atm

2= 430 L CO

More volume-volume:

2ZnS + 3O2 2SO2 ↑ + 2ZnO When 15 L of oxygen gas at 2.6 atm and 55°C are reacted as

shown, what volume of sulfur dioxide gas is produced at 65°C and 1.7 atm pressure?

More, volume-volume:

Plan of action: 2 2 2 2PV=nRT PV=nRTmoleratio

Vol O mol O mol SO Vol SO

2OPV = RT

n L atmmol K

2.6 atm 15 L=

0.0821 55+273K

2= 1.45 mol O 2

2

2 mol SO3 mol O

2= 0.966 mol SO

nRTV = P

2L atmmol K

0.966 mol H 0.0821 65+273K=

1.7 atm

2= 16 L SO