# Not so long ago, in a chemistry lab far far away… May the FORCE/area be with you.

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Not so long ago, in a chemistry lab far far away… May the FORCE/area be with you

1.Use the Ideal Gas Law to solve a gas stoichiometry problem.

When you know the guy has finally had enough!!!

2 H 2 O 2 (l) ---> 2 H 2 O (g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the volume of O 2 at STP? Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

2 H 2 O 2 (l) ---> 2 H 2 O (g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the volume of O 2 at STP? Solution 1.1 g H 2 O 2 1 mol H 2 O 2 1 mol O 2 22.4 L O 2 34 g H 2 O 2 2 mol H 2 O 2 1 mol O 2 34 g H 2 O 2 2 mol H 2 O 2 1 mol O 2 = 0.36 L O 2 at STP

How many grams of He are present in 8.0 L of gas at STP? = 1.4 g He 8.0 L He x 1 mol He 22.4 L He x 4.00 g He 1 mol He

Gas Stoichiometry Trick Page 41:1 If reactants and products are at the same conditions of temperature and pressure, then mole ratios of gases are also volume ratios. 3 H 2 (g) + N 2 (g)  2NH 3 (g) 3 moles H 2 + 1 mole N 2  2 moles NH 3 67.2 liters H 2 + 22.4 liter N 2  44.8 liters NH 3

Gas Stoichiometry Trick Example How many liters of ammonia can be produced when 12 liters of hydrogen react with an excess of nitrogen in a closed container at constant temperature? 3 H 2 (g) + N 2 (g)  2NH 3 (g) 12 L H 2 L H 2 = L NH 3 L NH 3 3 2 8.00

How to Report Computer Problems:

 1. You will need to use PV = nRT

Gas Stoichiometry Example Page x:1 How many liters of oxygen gas, at 1.00 atm and 25 o C, can be collected from the complete decomposition of 10.5 grams of potassium chlorate? 2 KClO 3 (s)  2 KCl(s) + 3 O 2 (g) 10.5 g KClO 3 1 mol KClO 3 122.55 g KClO 3 3 mol O 2 2 mol KClO 3 0.13 mol O 2

Gas Stoichiometry Example Page x:1 How many liters of oxygen gas, at 1.00 atm and 25 o C, can be collected from the complete decomposition of 10.5 grams of potassium chlorate? 2 KClO 3 (s)  2 KCl(s) + 3 O 2 (g) 3.2 L O 2 (1.0 atm)(V)(0.08206 atm*L / mol*K )(298 K)(0.13 mol)

1.Use the Ideal Gas Law to solve a gas stoichiometry problem.

Now we’re in big Do-Do

Gas Stoichiometry Page x:2 1 P 4 (s) + 6 H 2 (g)  4 PH 3 (g) 2.51 g P 4 1 mol P 4 123.88 g P 4 6 mol H 2 1 mol P 4 0.122 mol H 2

3.0 L H 2 (0.991 atm)(V)(0.08206 atm*L / mol*K )(298 K)(0.122 mol) Gas Stoichiometry Page x:2 1 P 4 (s) + 6 H 2 (g)  4 PH 3 (g)

Quicklime Example Page x:3 1 CaCO 3 (s)  1 CaO (s) + 1 CO 2 (g) 152 g CaCO 3 100.09 g 1 mol CaCO 3 1 mol CO 2 1 mol CaCO 3 = 34.0 L CO 2 at STP 1 mol CO 2 22.4 L CO 2

Pass the Clicker!!!

1. ZnCl + H 2. ZnCl + H 2 3. Zn 2 Cl + H 2 4. ZnCl 2 + H 2 5. Not listed

1. DD 2. SD 3. Synthesis 4. Decomposition 5. Not listed

1. 1,2,1,2 2. 2,1,2,1 3. 1,1,1,1 4. 2,2,1,2 5. Not listed

1. 250 atm 2. 314 atm 3. 0.329 atm 4. 0.987 atm 5. Not listed

1. 0.112 g 2. 1.18 g 3. 628 g 4. 0.000 628 g 5. 0.369 g

1.Explain Dalton’s Law and use it to solve a problem. 2.Use the Ideal Gas Law to solve a gas density problem. 3.Use the Ideal Gas Law to solve a gas stoichiometry problem.

Highdensity Lowerdensity 22.4 L of ANY gas AT STP = 1 mole

Gas Density … so at STP…

Density and the Ideal Gas Law Combining the formula for density with the Ideal Gas law, substituting and rearranging algebraically: M = Molar Mass P = Pressure R = Gas Constant T = Temperature in Kelvins

Gas Stoichiometry #4 How many liters of oxygen gas, at 37.0  C and 0.930 atmospheres, can be collected from the complete decomposition of 50.0 grams of potassium chlorate? 2 KClO 3 (s)  2 KCl(s) + 3 O 2 (g) = “n” mol O 2 50.0 g KClO 3 1 mol KClO 3 122.55 g KClO 3 3 mol O 2 2 mol KClO 3 = 0.612 mol O 2 = 16.7 L

How many L of O 2 are needed to react 28.0 g NH 3 at 24°C and 0.950 atm? 4 NH 3 (g) + 5 O 2 (g) 4 NO(g) + 6 H 2 O(g)

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