wrap up proving “r” constant. we can find out the volume of gas through stoichiometry ch 4 + o...
TRANSCRIPT
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