ideal gas equation pv=nrt practice problems. 2 alcl 3 2 al + 3 cl 2 ↑ if 13 g of aluminum...
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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