Announcements. 1 point for every question attempted; 0.2 extra credit points for every correct answer

Transcription

1 Announcements Print worksheet #4 prior to your Thursday discussion section A full schedule of readings and suggested problems is posted on the course website LON-CAPA assignments 1 and 2 are due Friday at 9am Don t forget to bring your Clicker to class EVERY day 1 point for every question attempted; 0.2 extra credit points for every correct answer

2 Chemical Reactions Hydrogen and oxygen gases mix to form liquid water The starting materials in a reaction are known as the reactants The end results are known as the products Hydrogen (white) and oxygen (red) are naturally found as diatomics + 2 H atoms 2 H atoms 2 O atoms 1 O atom Chemical reactions can rearrange atoms, but cannot create or destroy them. Something must be wrong with this picture!

3 Chemical Reactions The reaction must be balanced so that there are equal atoms of each type on both sides of the equation (Conservation of Mass) + 4 H atoms 4 H atoms 2 O atoms 2O atoms The balanced chemical reaction looks like 2H 2 + O 2 2H 2 O (g) (g) (l) (s) = solid (aq) = aqueous Δ = heat hν = light

4 coefficients H 2 + O 2 H 2 O 2 2 (g) (g) (l) atomic 4 H atoms 2 O atoms 4 H atoms 2 O atoms molecular 2 H 2 1 O 2 2 H 2 O molar 2 mol H 2 1 mol O 2 2 mol H 2 O mass 2 x g H g O 2 2 x18.02 g H 2 O g reactants g products Stoichiometry tells us the ratio between each reactant and product in a balanced equation

5 Solid iron reacts with oxygen to form solid iron(iii) oxide Determine reactants and products: Fe (s) O 2 (g) Fe 2 O 3 (s) Write the unbalanced equation: Fe (s) + O 2 (g) Fe 2 O 3 (s) 1 Fe atom, 2 oxygen atoms 2 Fe atoms, 3 oxygen atoms Balance the atoms of one element: 2Fe (s) + O 2 (g) Fe 2 O 3 (s) 2 Fe atoms, 2 oxygen atoms 2 Fe atoms, 3 oxygen atoms

6 Solid iron reacts with oxygen to form solid iron(iii) oxide Balance the atoms of another element: 2Fe (s) + 3O 2 (g) 2Fe 2 O 3 (s) 2 Fe atoms, 6 oxygen atoms 4 Fe atoms, 6 oxygen atoms Make corrections as needed until all elements are balanced: 4Fe (s) + 3O 2 (g) 2Fe 2 O 3 (s) 4 Fe atoms, 6 oxygen atoms 4 Fe atoms, 6 oxygen atoms Finally all the atoms are balanced. Mass must also be conserved: 4 (55.84 amu) + 3 (32.00 amu) 2 ( amu) amu amu

7 Magnesium reacts with hydrochloric acid to make hydrogen gas and magnesium chloride Determine reactants and products: Mg (s) HCl (aq) H 2 (g) MgCl 2 (aq) Write the unbalanced equation: Mg (s) + HCl (aq) H 2 (g) + MgCl 2 (aq) 1 Mg atom, 1 H atom, 1 Cl atom 1 Mg atom, 2 H atoms, 2 Cl atoms Balance the equation: Mg (s) + 2HCl (aq) H 2 (g) + MgCl 2 (aq) 1 Mg atom, 2 H atoms, 2 Cl atoms 1 Mg atom, 2 H atoms, 2 Cl atoms

8 Ammonia (gas) reacts with oxygen to produce nitrogen monoxide (gas) and water Determine reactants and products: NH 3 (g) O 2 (g) NO (g) H 2 O (l) Write and balance the equation. What is the coefficient for NH 3 in the balanced equation? A. 1 B. 2 C. 3 D. 4 E. 5 Use your clickers to select the correct response

9 If we start with 10.0 g of ammonia (and unlimited oxygen), how many moles of water (l) can we produce? 4NH 3 (g) + 5 O 2 (g) 4NO (g) + 6H 2 O (l) 10.0 g NH 3 1 mol NH g NH 3 = mol NH mol NH 3 6 mol H 2 O 4 mol NH 3 = mol H 2 O Use the coefficients of the balanced equation to convert between moles of any two reactants or products If we start with 10.0 g of oxygen (and unlimited ammonia), how many moles of water (l) can we produce? 10.0 g O 2 1 mol O 2 6 mol H 2 O 32.0 g O = mol H2 O 2 5 mol O 2

10 If we start with 10.0 g of ammonia and 10.0 g of oxygen, how many moles of water (l) can we produce? 4NH 3 (g) + 5 O 2 (g) 4NO (g) + 6H 2 O (l) With 10.0 g of ammonia, we could produce mol H 2 O With 10.0 g of oxygen, we could produce mol H 2 O In this case, oxygen is the limiting reagent, the reactant that limits the amount of product that can be produced The theoretical yield is the maximum amount of product that could be formed in a reaction mol H 2 O 18.0 g H 2 O = 6.75 g H2 O 1 mol H 2 O

11 Which reagent will be left after the reaction is complete? How much of it will remain (g)? 4NH 3 (g) + 5 O 2 (g) 4NO (g) + 6H 2 O (l) Oxygen is the limiting reagent, so it will be used completely during the reaction. Some amount of ammonia (NH 3 ) will be left after the reaction is complete. To determine the amount of excess ammonia, first determine how much is used during the reaction: mol H 2 O 4 mol NH g NH 3 = 4.25 g NH 6 mol H 2 O 3 consumed 1 mol NH 3 Excess = 10.0 g NH g NH 3 = 5.75 g NH 3

12 Percent Yield 4NH 3 (g) + 5 O 2 (g) 4NO (g) + 6H 2 O (l) We calculated that 6.75 g H 2 O could be produced from a mixture containing 10.0 g of each reactant In real reactions, the actual amount produced will always be smaller than the theoretical yield The percent yield is a measure of how close the actual yield is to the theoretical yield % yield = actual yield x 100% theoretical yield Suppose only 5.25 g of water were produced in the previous reaction. What is the percent yield? % yield = 5.25 g H 2 O x 100% = 77.8% 6.75 g H 2 O