Transcript NOTES: Gases, Molar Volume, & Stoichiometry – a REVIEW!

NOTES: Gases, Molar Volume,
& Stoichiometry – a
REVIEW!
Avogadro’s Hypothesis:
● equal volumes of gases at the same
temperature and pressure contain equal
#’s of particles;
● at STP, 1 mol (6.02 x 1023) of particles of
any gas occupies a volume of 22.4 L.
Example #1:
What volume does 0.742 mol of argon
gas occupy at STP?
Example #1:
What volume does 0.742 mol of argon
gas occupy at STP?
0.742 mol Ar
22.4 L

1 mol Ar
Example #1:
What volume does 0.742 mol of argon
gas occupy at STP?
0.742 mol Ar
22.4 L

1 mol Ar
= 16.6 L
Example #2:
How many oxygen molecules are in 3.36 L of
oxygen gas at STP?
Example #2:
How many oxygen molecules are in 3.36 L of
oxygen gas at STP?

23 


6
.
02
x
10
 1mol O2  


3.36L O2 
 22.4 L   1mol O 2 


Example #2:
How many oxygen molecules are in 3.36 L of
oxygen gas at STP?

23 
 1mol O2   6.02 x10  = 9.03 x 1022

3.36L O2 
molecules


22
.
4
L
1
mol
O
2





Example #3:
Determine the volume (in L) occupied by 12.6 g
of nitrogen gas, N2, at STP.
Example #3:
Determine the volume (in L) occupied by 12.6 g
of nitrogen gas, N2, at STP.
 1mol N2   22.4L N 2 
 

12.6 g 
 28.0 g   1mol N2 
Example #3:
Determine the volume (in L) occupied by 12.6 g
of nitrogen gas, N2, at STP.
 1mol N2   22.4L N 2 
 

12.6 g 
 28.0 g   1mol N2 
= 10.0 L N2
Stoichiometry Example #1:
Li3N(g) + 3H2O(l)  NH3(g) + 3LiOH(aq)
What mass of water is needed to react with 29.3
L of Li3N?
Stoichiometry Example #1:
Li3N(g) + 3H2O(l)  NH3(g) + 3LiOH(aq)
What mass of water is needed to react with 29.3
L of Li3N?
 1mol Li3N   3molH2O  18.0 g 
 


29.3L 
 22.4 L   1mol Li3N  1mol H2O 
Stoichiometry Example #1:
Li3N(g) + 3H2O(l)  NH3(g) + 3LiOH(aq)
What mass of water is needed to react with 29.3
L of Li3N?
 1mol Li3N   3molH2O  18.0 g 
 


29.3L 
 22.4 L   1mol Li3N  1mol H2O 
= 70.6 g H2O
Stoichiometry Example #2:
Li3N(g) + 3H2O(l)  NH3(g) + 3LiOH(aq)
When 13.3 L of NH3 are produced how many
formula units of LiOH are produced?
Stoichiometry Example #2:
Li3N(g) + 3H2O(l)  NH3(g) + 3LiOH(aq)
When 13.3 L of NH3 are produced how many
formula units of LiOH are produced?
 1mol NH3   3molLiOH  6.02x10 f.u. 

 

13.3L 
 22.4 L   1mol NH3  1mol LiOH 
23
Stoichiometry Example #2:
Li3N(g) + 3H2O(l)  NH3(g) + 3LiOH(aq)
When 13.3 L of NH3 are produced how many
formula units of LiOH are produced?
 1mol NH3   3molLiOH  6.02x10 f.u. 

 

13.3L 
 22.4 L   1mol NH3  1mol LiOH 
23
= 1.07 x 1023 formula
units LiOH
Stoichiometry Example #3:
Li3N(g) + 3H2O(l)  NH3(g) + 3LiOH(aq)
Given 112.5 grams of LiOH produced, what
volume (L) of Li3N was used?
Stoichiometry Example #3:
Li3N(g) + 3H2O(l)  NH3(g) + 3LiOH(aq)
Given 112.5 grams of LiOH produced, what
volume (L) of Li3N was used?
 1mol LiOH   1molLi3N  22.4 L 
 


112.5 gLiOH 
 23.9 g   3 mol LiOH  1mol Li3N 
Stoichiometry Example #3:
Li3N(g) + 3H2O(l)  NH3(g) + 3LiOH(aq)
Given 112.5 grams of LiOH produced, what
volume (L) of Li3N was used?
 1mol LiOH   1molLi3N  22.4 L 
 


112.5 gLiOH 
 23.9 g   3 mol LiOH  1mol Li3N 
= 35.1 L Li3N