Wrap up Proving “R” constantEntry Task: April 18th-19th B.2

We can find out the volume of gas through Stoichiometry

CH4 + O2 CO2 + H2O22

Think of the coefficients as volume ratios

1 liter (volume) of CH4

2 liters (volume) of O2

1 liter (volume) of CO2

2 liters (volume) of H2O

CH4 + O2 CO2 + H2O22

What if I have 3.5 liters of CH4, how much oxygen is needed for this reaction?

3.5 L of CH4

1 liter of CH4

2 liters of O2= 7 liters of O2

CH4 + O2 CO2 + H2O22

If 4.00L of oxygen gas reacts completely by this reaction at a constant pressure and temperature of 2.00 atm and 300K, how many grams of water are produced?

P = VO2 = T=

mH2O = R =

2.00 atm 4.00L O2 300K

X 0.0821

ALL data is based off of Oxygen gas

CH4 + O2 CO2 + H2O22P = VO2 = T=

mH2O = R =

2.00 atm 4.00L O2 300K

X 0.0821

Use the stoich- relationship between O2 and Water!

4.00 L of O2

2 liter of O2

2 liters of H2O= 4.00 liters of H2O

CH4 + O2 CO2 + H2O22

P = VH2O = T=

mH2O = R =

2.00 atm 4.00L H2O 300K

X 0.0821

NOW we have a volume (4.00L of Water), next we can plug it into the ideal gas law.

**noticed the change in volumes label!!

PV = nRT

(2.00 atm)(4.00 L) = (X H20)(0.0821)(300K)

P = VH2O = T=

mH2O = R =

2.00 atm 4.00L H2O 300K

X 0.0821

PV = nRT

(2.00 atm)(4.0 L)

(0.0821)(300 K) = X mol

(2.00 atm)(4.00 L) = (X H20)(0.0821)(300K)

DO the MATH

8

24.63 = 0.32 mol H2O

(2.00 atm)(4.0 L)

(0.0821)(300 K) = X mol

0.32 mol H2O

Not done yet!!! We have to convert this to grams of H2O

0.32 mol H2O

1 of mol H2O

18 g of H2O= 5.85 g of H2O

N2 + H2 NH3 23

Ammonia is synthesized from hydrogen and nitrogen gas.

If 5.00L of nitrogen reacts completely by this reaction at a constant pressure and temperature of 3.00 atm and 298K, how many grams of ammonia are produced?

N2 + H2 NH3 23

If 5.00L of nitrogen reacts completely by this reaction at a constant pressure and temperature of 3.00 atm and 298K, how many grams of ammonia are produced?

VN2 = 5.00L

P = 3.00 atm

T = 298K

mNH3 = X

R = 0.0821

ALL data is based off of Nitrogen gas

N2 + H2 NH3 23

VN2 = 5.00L

P = 3.00 atm

T = 298K

mNH3 = X

R = 0.0821

5.0 L of N2

1 liter of N2

2 liters of NH3= 10 liters of NH3

Use the stoich- relationship between H2 and NH3!

N2 + H2 NH3 23

NOW we have a volume (10L of NH3), next we can plug it into the ideal gas law.

VNH3 = 10 L

P = 3.00 atm

T1 = 298K

mNH3 = X

R = 0.0821

PV = nRT

VNH3 = 10. L

P = 3.00 atm

T1 = 298K

mNH3 = X

R= 0.0821

(3.00 atm)(10 L) = (X NH3)(0.0821 )(298K)

PV = nRT

(3.00 atm)(10 L) = (X NH3)(0.0821 )(298K)

(3.00 atm)(10 L)

(0.0821 ) (298 K) = X mol

DO the MATH

30

24.47 = 1.23 mol NH3

= X mol(3.00)(10 )

(0.0821 mol)(298)

Are we done??

1.23 mol NH3

No!! We have to convert this to grams of NH3

1.23 mol NH3

1 mol of NH3

17.04 g of NH3

= 21.0 g of NH3

1. Balance gas equation2. From balancing – get gas ratio3. Use given volume to put into volume ratio to get

unknowns’ volume

4. Plug numbers into ideal gas law equation5. Get X by itself and do the math*6. If needed, convert mole to grams

Summary on Gas Stoich problems

Ammonia nitrate is a common ingredient in a chemical fertilizer. Use the reaction shown to calculate the mass of solid ammonium nitrate that must be used to obtain 0.100 L of dinitrogen monoxide gas at STP

NH4NO3 N2O + H2O2

There is 0.100 liters of N2O, how much oxygen is needed for this reaction

0.100 L of N2O

1 liter of N2O

1 liters of NH4NO3

= 0.100 L of NH4NO3

NH4NO3 N2O + H2O2

PV = nRT

VNH4NO3 = 0.100 L

P = 1.00 atm

T1 = 0.00 + 273 = 273K

mNH4NO3 = X

R= 0.0821

(1.00 atm)(0.100 LNH4NO3) =(X NH4NO3)(0.0821)(273K)

Ammonia nitrate is a common ingredient in a chemical fertilizer. Use the reaction shown to calculate the mass of solid ammonium nitrate that must be used to obtain 0.100 L of dinitrogen oxide gas at STP

PV = nRT

(1.00 atm)(0.100 L) = (XNH4NO3)(0.0821)(273K)

(1.00 atm)(0.100 L)

(0.0821) (273K) = X molNH4NO3

DO the MATH

0.1

22.41 = 0.00446 mol NH4NO3

= X molNH4NO3

(1.00)(0.100 )

(0.0821)(273)

Are we done??= 0.00446 mol NH4NO3

No!! We have to convert this to grams of NH4NO3

0.00446 mol NH4NO3

1 mol of NH4NO3

80.04 g of NH4NO3

= 0.357 g of NH4NO3