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Chemistry: Atoms First Julia Burdge & Jason Overby Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 8.3 and 8.4 Book Homework: 25, 27, 29, 31, 33, 37, 43, 47, 49, 51, 53, 55 and 57 Kent L. McCorkle Cosumnes River College Sacramento, CA
How to “Read” Chemical Equations 2 Mg + O 2 2 MgO 2 atoms Mg + 1 molecule O 2 makes 2 formula units MgO 2 moles Mg + 1 mole O 2 makes 2 moles MgO 48.6 grams Mg grams O 2 makes 80.6 g MgO IS NOT 2 grams Mg + 1 gram O 2 makes 2 g MgO 3.7
3 ways of representing the reaction of H 2 with O 2 to form H 2 O A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction reactantsproducts
Chemical Equations A chemical equation uses chemical symbols to denote what occurs in a chemical reaction. NH 3 + HCl → NH 4 Cl Ammonia and hydrogen chloride react to produce ammonium chloride. Each chemical species that appears to the left of the arrow is called a reactant. NH 3 + HCl → NH 4 Cl Each species that appears to the right of the arrow is called a product. NH 3 + HCl → NH 4 Cl 8.1
Interpreting and Writing Chemical Equations Labels are used to indicate the physical state: (g) gas (l) liquid (s) solid (aq) aqueous [dissolved in water] NH 3 (g) + HCl(g) → NH 4 Cl(s) SO 3 (g) + H 2 O(l) → H 2 SO 4 (aq)
Chemical equations must be balanced so that the law of conservation of mass is obeyed. Balancing is achieved by writing stoichiometric coefficients to the left of the chemical formulas. Balancing Chemical Equations
Generally, it will facilitate the balancing process if you do the following: 1)Change the coefficients of compounds before changing the coefficients of elements. 2)Treat polyatomic ions that appear on both sides of the equation as units. 3)Count atoms and/or polyatomic ions carefully, and track their numbers each time you change a coefficient.
Calculations with Balanced Chemical Equations Balanced chemical equations are used to predict how much product will form from a given amount of reactant. 2 moles of CO combine with 1 mole of O 2 to produce 2 moles of CO 2. 2 moles of CO is stoichiometrically equivalent to 2 moles of CO
Calculations with Balanced Chemical Equations Consider the complete reaction of 3.82 moles of CO to form CO 2. Calculate the number of moles of CO 2 produced.
Calculations with Balanced Chemical Equations Consider the complete reaction of 3.82 moles of CO to form CO 2. Calculate the number of moles of O 2 needed.
1.Write balanced chemical equation 2.Convert quantities of known substances into moles 3.Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4.Convert moles of sought quantity into desired units Mass Changes in Chemical Reactions 3.8
Stoichiometric Ratio Coefficient B Coefficient A Moles B = X Moles A The stoichiometric ratio is used to determine amounts of compounds consumed or produced in a balanced chemical reaction
Worked Example 8.5 Strategy Use the balanced chemical equation to determine the correct stoichiometric conversion factors, and then multiply by the number of moles of ammonia given. Urea [(NH 2 ) 2 CO] is a by-product of protein metabolism. This waste product is formed in the liver and then filtered from the blood and excreted in the urine by the kidneys. Urea can be synthesized in the laboratory by the combination of ammonia and carbon dioxide according to the equation (a) Calculate the amount of urea that will be produced by the complete reaction of 5.25 moles of ammonia. (b) Determine the stoichiometric amount of carbon dioxide required to react with 5.25 moles of ammonia.
Worked Example 8.5 (cont.) Solution (a) moles (NH 2 ) 2 CO produced = 5.25 mol NH 3 × (b) moles CO 2 produced = 5.25 mol NH 3 × 1 mol (NH 2 ) 2 CO 2 mol NH 3 = 2.63 mol (NH 2 ) 2 CO 1 mol CO 2 2 mol NH 3 = 2.63 mol CO 2 Think About It As always, check to be sure that units cancel properly in the calculation. Also, the balanced equation indicates that there will be fewer moles of urea produced than ammonia consumed. Therefore, your calculated number of moles of urea (2.63) should be smaller than the number of moles given in the problem (5.25). Similarly, the stoichiometric coefficients in the balanced equation are the same for carbon dioxide and urea, so your answers to this problem should also be the same for both species.
Worked Example 8.6 Strategy For part (a), use the molar mass of nitrous oxide to convert the given mass of nitrous oxide to moles, use the appropriate stoichiometric conversion factor to convert to moles of ammonium nitrate, and then use the molar mass of ammonium nitrate to convert to grams of ammonium nitrate. For part (b), use the molar mass of nitrous oxide to convert the given mass of nitrous oxide to moles, use the stoichiometric conversion factor to convert from moles of nitrous oxide to moles of water, and then use the molar mass of water to convert to grams of water. Dinitrogen monoxide (N 2 O), also known as nitrous oxide or “laughing gas,” is used as an anesthetic in dentistry. It is manufactured by heating ammonium nitrate. The balanced equation is NH 4 NO 3 (s) → N 2 O(g) + 2H 2 O(l) (a) Calculate the mass of ammonium nitrate that must be heated in order to produce 10.0 g of nitrous oxide. (b) Determine the corresponding mass of water produced in the reaction. Δ
Worked Example 8.6 (cont.) Solution (a) 10.0 g N 2 O × mol N 2 O × mol NH 4 NO 3 × Thus, 18.2 g of ammonium nitrate must be heated in order to produce 10.0 g of nitrous oxide. (b) Starting with the number of moles of nitrous oxide determined in the first step of (a), mol N 2 O × mol H 2 O × 1 mol N 2 O g N 2 O = mol N 2 O 1 mol NH 4 NO 3 1 mol N 2 O = mol NH 4 NO g NH 4 NO 3 1 mol NH 4 NO 3 = 18.2 g NH 4 NO 3 2 mol H 2 O 1 mol N 2 O = mol H 2 O g H 2 O 1 mol H 2 O = 8.18 g H 2 O Think About It Use the law of conservation of mass to check your answers. Make sure that the combined mass of both products is equal to the mass of reactant you determined in part (a). In this case (rounded to the appropriate number of significant figures), 10.0 g g = 18.2 g. Remember that small differences may arise as the result of rounding.
The reactant used up first in a reaction is called the limiting reactant. Excess reactants are those present in quantities greater than necessary to react with the quantity of the limiting reactant. CO(g) + 2H 2 (g) → CH 3 OH(l) Limiting Reactants 8.4
Limiting Reactants Consider the reaction between 5 moles of CO and 8 moles of H 2 to produce methanol. How many moles of H 2 are necessary in order for all the CO to react? How many moles of CO are necessary in order for all of the H 2 to react? 10 moles of H 2 required; 8 moles of H 2 available; limiting reactant. 4 moles of CO required; 5 moles of CO available; excess reactant. CO(g) + 2H 2 (g) → CH 3 OH(l)
Worked Example 8.7 Strategy Convert each of the reactant masses to moles. Use the balanced equation to write the stoichiometric conversion factor and determine which reactant is limiting. Next, determine the number of moles of excess reactant remaining and the number of moles of CO 2 produced. Finally, use the appropriate molar masses to convert moles of excess reactant and moles of CO 2 to grams. Alka-Seltzer tablets contain aspirin, sodium bicarbonate, and citric acid. When they come into contact with water, the sodium bicarbonate (NaHCO 3 ) and citric acid (H 3 C 6 H 5 O 7 ) react to form carbon dioxide gas, among other products. 3NaHCO 3 (aq) + H 3 C 6 H 5 O 7 (aq) → 3CO 2 (g) + 3H 2 O(l) + Na 3 C 6 H 5 O 7 (aq) The formation of CO 2 causes the trademark fizzing when the tablets are dropped into a glass of water. An Alka-Seltzer tablet contains g of sodium bicarbonate and g citric acid. Determine, for a single tablet dissolved in water, (a) which ingredient is the limiting reactant, (b) what mass of the excess reactant is left over when the reaction is complete, and (c) what mass of CO 2 forms.
Worked Example 8.7 (cont.) Solution g NaHCO 3 × g H 3 C 6 H 5 O 7 × (a) To determine which reactant is limiting, calculate the amount of citric acid necessary to react completely with mol sodium bicarbonate mol NaHCO 3 × The amount of H 3 C 6 H 5 O 7 required to react with mol of NaHCO 3 is more than a tablet contains. Therefore, citric acid is the limiting reactant and sodium bicarbonate is the excess reactant. 1 mol NaHCO g NaHCO 3 = mol NaHCO 3 1 mol H 3 C 6 H 5 O g H 3 C 6 H 5 O 7 = mol H 3 C 6 H 5 O 7 1 mol H 3 C 6 H 5 O 7 3 mol NaHCO 3 = mol H 3 C 6 H 5 O 7
Worked Example 8.7 (cont.) Solution (b) To determine the mass of excess reactant (NaHCO 3 ) left over, first calculate the amount of NaHCO 3 that will react: mol H 3 C 6 H 5 O 7 × Thus, mole of NaHCO 3 will be consumed, leaving mole unreacted. Convert the unreacted amount to grams as follows: mol NaHCO 3 × g NaHCO 3 1 mol NaHCO 3 = g NaHCO 3 3 mol NaHCO 3 1 mol H 3 C 6 H 5 O 7 = mol NaHCO 3
Worked Example 8.7 (cont.) Solution (c) To determine the mass of CO 2 produced, first calculate the moles of CO 2 produced from the number of moles of limiting reactant (H 3 C 6 H 5 O 7 ) consumed mol H 3 C 6 H 5 O 7 × Convert this amount to grams as follows: mol CO 2 × To summarize the results: (a) citric acid is the limiting reactant, (b) g sodium bicarbonate remains unreacted, and (c) g carbon dioxide is produced g CO 2 1 mol CO 2 = g CO 2 3 mol CO 2 1 mol H 3 C 6 H 5 O 7 = mol CO 2 Think About It In a problem such as this, it is a good idea to check your work by calculating the amounts of the other products in the reaction. According to the law of conservation of mass, the combined starting mass of the two reactants (1.700 g g = g) should equal the sum of the masses of products and leftover excess reactant. In this case, the masses of H 2 O and Na 3 C 6 H 5 O 7 produced are g and g, respectively. The mass of CO 2 produced is g [from part (c)] and the amount of excess NaHCO 3 is g [from part (b)]. The total, g g g g, is g, identical to the total mass of reactants.
Limiting Reagent The number of bicycles that can be assembled is limited by whichever part runs out first. In the inventory shown in this figure, wheels are that part. The reactant that is completely consumed by the reaction
Limiting Reactants The theoretical yield is the amount of product that forms when all the limiting reactant reacts to form the desired product. The actual yield is the amount of product actually obtained from a reaction. The percent yield tells what percentage the actual yield is of the theoretical yield.
Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield x
Worked Example 8.8 Strategy Convert reactant grams to moles, and determine which is the limiting reactant. Use the balanced equation to determine the moles of aspirin that can be produced and convert to grams for the theoretical yield. Use this and the actual yield given to calculate the percent yield. Aspirin, acetylsalicylic acid (C 9 H 8 O 4 ), is the most commonly used pain reliever in the world. It is produced by the reaction of salicylic acid (C 7 H 6 O 3 ) and acetic anhydride (C 4 H 6 O 3 ) according to the following equation: In a certain aspirin synthesis, g of salicylic acid and g of acetic anhydride are combined. Calculate the percent yield if g of aspirin are produced. C 7 H 6 O 3 salicylic acid C 4 H 6 O 3 acetic anhydride C 9 H 8 O 4 acetylsalicylic acid HC 2 H 3 O 2 acetic acid ++→
Worked Example 8.8 (cont.) Solution g C 7 H 6 O 3 × g C 4 H 6 O 3 × Because the two reactants combine in a 1:1 mole ratio, the reactant present in the smallest number of moles (in this case, salicylic acid) is the limiting reactant. According to the balanced equation, one mole of aspirin is produced for every mole of salicylic acid consumed. Therefore, the theoretical yield of aspirin is mol. We convert this to grams using the molar mass of aspirin: mol C 9 H 8 O 4 × 1 mol C 7 H 6 O g C 7 H 6 O 3 = mol C 7 H 6 O 3 1 mol C 4 H 6 O g C 4 H 6 O 3 = mol C 4 H 6 O g C 9 H 8 O 4 1 mol C 9 H 8 O 4 = g C 9 H 8 O 4
Methanol burns in air according to the equation 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH 3 OHmoles CH 3 OHmoles H 2 Ograms H 2 O molar mass CH 3 OH coefficients chemical equation molar mass H 2 O 209 g CH 3 OH 1 mol CH 3 OH 32.0 g CH 3 OH x 4 mol H 2 O 2 mol CH 3 OH x 18.0 g H 2 O 1 mol H 2 O x = 235 g H 2 O 3.8