Basic Chemistry Chapter 11 Gases Chapter 11 Lecture Fourth Edition Chapter 11 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.

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

Guide to Reactions Involving the Ideal Gas Law

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)

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 + 273 = 300 K)

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)

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.

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.

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.