1. Basic Chemistry Chapter 11 Gases Chapter 11 Lecture
Fourth Edition
Chapter Gases
11.9 Gas Laws and Chemical Reactions
Learning Goal Determine the mass or volume of a gas that reacts or forms in a chemical reaction.

2. Gases in Equations
The volume or amount of a gas in a chemical reaction can be calculated from
the ideal gas law
mole–mole factors from the balanced equation
molar mass

3. Guide to Reactions Involving the Ideal Gas Law

4. Learning Check
What volume (L) of Cl2 gas at 1.20 atm and 27 °C is needed to completely react with 1.50 g of Al?
2Al(s) + 3Cl2(g)  2AlCl3(s)

5. Step 1 State the given and needed quantities.
Solution
What volume (L) of Cl2 gas at 1.20 atm and 27 °C is needed to completely react with 1.50 g of Al?
Step 1 State the given and needed quantities.
Equation: 2Al(s) + 3Cl2(g)  2AlCl3(s)
Given
Need
Reactant:
liters of Cl2(g)
Product: Cl2(g) at 1.20 atm,
27 oC (27 oC = 300 K)

6. Step 2 State the given and needed quantities.
Solution
What volume (L) of Cl2 gas at 1.20 atm and 27 °C is needed to completely react with 1.50 g of Al?
2Al(s) + 3Cl2(g)  2AlCl3(s)
Step 2 State the given and needed quantities.
Molar mass
Mole-mole factor
grams
of Al
moles
of Al
moles of
Cl2(g)

7. Solution
What volume (L) of Cl2 gas at 1.20 atm and 27 °C is needed to completely react with 1.50 g of Al?
2Al(s) + 3Cl2(g)  2AlCl3(s)
Step 3 Write the equalities and conversion factors for molar mass and mole–mole factors.

8. Solution
What volume (L) of Cl2 gas at 1.20 atm and 27 °C is needed to completely react with 1.50 g of Al?
2Al(s) + 3Cl2(g)  2AlCl3(s)
Step 4 Set up the problem to calculate moles of needed quantity.

9. Solution
What volume (L) of Cl2 gas at 1.20 atm and 27 °C is needed to completely react with 1.50 g of Al?
2Al(s) + 3Cl2(g)  2AlCl3(s)
Step 5 Convert the moles of needed quantity to mass or volume using the molar mass or the ideal gas law equation.