1. PVm = nRT Molar Volume: Vm = _____ 22.4 L 1 mol the ______ of ______
of any gas at ____.
volume
1 mole
STP
PVm = nRT
(1.00 atm) Vm =
(1 mol )( )(273 K)
The volume of 1 mole of any gas at STP will be:
Vm = _____
22.4 L
1 mol
but…
ONLY at STP!!!

2. Gas Stoich with Molar Volume (22.4 L/mol at STP)
L  L
What volume of NH3(g) can be produced from reacting completely 5.60 L of N2(g) at STP?
3 H2(g) + N2(g)  2 NH3(g) 1
mol N2 2
mol NH3
22.4
L NH3
5.60 L N2 x x x
22.4
L N2 1
mol N2 1
mol NH3 2
mol NH3 OR
5.60 L N2 x
= ____L NH3
11.2 L NH3 1
mol N2

3. Gas Stoich with Molar Volume (22.4 L/mol at STP)
L  g
Calculate the mass of H2(g) consumed to produce 44.8 L of NH3(g) at STP.
3 H2(g) + N2(g)  2 NH3(g) 1
mol NH3 3
mol H2
2.02
g H2
44.8 L NH3 x x x
22.4
L NH3 2
mol NH3 1
mol H2
= ____g H2
6.06 g H2

4. Now you try this one alone:
Gas Stoich with Molar Volume (22.4 L/mol at STP)
Now you try this one alone:
g  L
What volume of O2 gas is produced when 980 g KClO3 is decomposed at STP?
KClO3(s)  KClO(s) + O2(g) 1
mol KClO3 1
mol O2
22.4
L O2
980 g KClO3 x x x
122.55
g KClO3 1
mol KClO3 1
mol O2
= ____L O2
179 L O2
Molar Mass of KClO3 is g/mol

5. KClO3(s)  KClO(s) + O2(g)
NOT at STP
PV = nRT
use…
What volume of O2 gas is produced from 490 g KClO3 at 298 K and 1.06 atm?
KClO3(s)  KClO(s) + O2(g) 1
mol KClO3 1
mol O2
490 g KClO3 x x =
4.00
mol O2
122.55
g KClO3 1
mol KClO3
(1.06 atm) V =
(4.00 mol )( )(298 K)
= ____L O2
92.3 L O2
Molar Mass of KClO3 is g/mol