1/7/013 Chemistry: Atoms First Julia Burdge & Jason Overby 8 Stoichiometry: Ratios of Combination Chapter 8 Chemical Reactions Kent L. McCorkle Cosumnes River College Sacramento, CA Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 8.1 Chemical Equations Interpreting and Writing Chemical Equations Patterns of Chemical Reactivity 8. Combustion Analysis Determination of Empirical Formula 8.3 Calculations with Balanced Chemical Equations Moles of Reactants and Products Mass of Reactants and Products 8.4 Limiting Reactants Determining the Limiting Reactant Reaction Yield 8.5 Periodic Trends in Reactivity of the Main Group Elements General Trends in Reactivity Reactions of the Active Metals Reactions of the Other Main Group Elements Comparison of Group 1A and Group 1B Elements 8.1 Chemical Equations Interpreting and Writing 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 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 O(l) H SO 4 (aq) Chemical equations must be balanced so that the law of conservation of mass is obeyed. Generally, it will facilitate the balancing process if you do the following: 1) Change the coefficients of compounds before changing the coefficients of elements. Balancing is achieved by writing stoichiometric coefficients to the left of the chemical formulas. ) 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. 1
1/7/013 Worked Example 8.1 Write the balanced chemical equation that represents the combustion of propane. : Step 1: Write the unbalanced equation: C 3 H 8 (g) + O (g) CO (g) + H O(l) Step : Leaving O until the end, balance each of the atoms: C 3 H 8 (g) + 5O (g) 3CO (g) + 4H O(l) Step 3: Double check to make sure there are equal numbers of each type on atom on both sides of the equation. Write and balance the chemical equation for the aqueous reaction of barium hydroxide and perchloric acid to produce aqueous barium perchlorate and water. Strategy The reactants are Ba(OH) and HClO 4, and the products are Ba(ClO 4 ) and H O. Because the reaction is aqueous, all species except H O will be labeled (aq) in the equation. Being a liquid, H O will be labeled (l). Adjust the coefficients to ensure that there are identical numbers of each type of atom on both sides of the reaction arrow. The chemical statement barium hydroxide and perchloric acid react to produce barium perchlorate and water can be represented with the following unbalanced equation: Ba(OH) (aq) + HClO 4 (aq) Ba(ClO 4 ) (aq) + H O(l) Perchlorate ions (ClO 4- ) appear on both sides of the equation, so count them as units, rather than count the individual atoms they contain. Thus, the tally of atoms and polyatomic ions is Worked Example 8.1 (cont.) (cont.) Ba(OH) (aq) + HClO 4 (aq) Ba(ClO 4 ) (aq) + H O(l) 1 Ba 1 O 1 (not including O atoms in ClO - 4 ions) 3 H 1 ClO - 4 The barium atoms are already balanced, and placing a coefficient of in front of HClO 4 (aq) balances the number of perchlorate ions. Ba(OH) (aq) + HClO 4 (aq) Ba(ClO 4 ) (aq) + H O(l) 1 Ba 1 O 1 (not including O atoms in ClO - 4 ions) 4 H ClO - 4 Worked Example 8.1 (cont.) (cont.) Placing a in front of H O(l) balances both the O and H atoms, giving us the final balanced equation: Ba(OH) (aq) + HClO 4 (aq) Ba(ClO 4 ) (aq) + H O(l) 1 Ba 1 O (not including O atoms in ClO - 4 ions) 4 H 4 ClO - 4 Think About It Check to be sure the equation is balanced by counting all the atoms individually. 1 Ba 1 10 O 10 4 H 4 Cl Worked Example 8. Butyric acid (also known as butanoic acid, C 4 H 8 O ) is one of many compounds found in milk fat. First isolated from rancid butter in 1869, burtyic acid has received a great deal of attention in recent years as a potential anticancer agent. Write a balanced equation for the metabolism of butyric acid. Assume that the overall process of metabolism and combustion are the same (i.e., reaction with oxygen to produce carbon dioxide and water). Strategy Begin by writing an unbalanced equation to represent the combination of reactants and formation of products as stated in the problem, and then balance the equation. Worked Example 8. (cont.) C 4 H 8 O (aq) + O (g) CO (g) + H O(l) Balance the number of C atoms by changing the coefficient for CO from 1 to 4. C 4 H 8 O (aq) + O (g) 4CO (g) + H O(l) Balance the number of H atoms by changing the coefficient for H O from 1 to 4. C 4 H 8 O (aq) + O (g) 4CO (g) + 4H O(l) Finally, balance the number of O atoms by changing the coefficient for O from 1 to 5. C 4 H 8 O (aq) + 5O (g) 4CO (g) + 4H O(l) Think About It Count the number of each type of atom on each side of the reaction arrow to verify that the equation is properly balanced. The are 4 C, 8 H, and 1 O in the reactants and in the products, so the equations is balanced.
1/7/013 Patterns of Chemical Reactivity Worked Example 8.3 Three of the most commonly encountered reaction types are combination, decomposition, and combustion. Combination two or more reactants combine to form a single product NH 3 (g) + HCl(g) NH 4 Cl(s) Decomposition two or more products form from a single reactant Δ CaCO 3 (s) CaO(s) + CO (g) Combustion a substance burns in the presence of oxygen. Combustion of a compound that contains C and H (or C, H, and O) produces carbon dioxide gas and water. CH O(l) + O (g) CO (g) + H O(l) Determine whether each of the following equations represents a combination reaction, a decomposition reaction, or a combustion reaction: (a) H (g) + Br (g) HBr(g), (b) HCO H(l) + O (g) CO (g) + H O(l), (c) KClO 3 (s) KCl(s) + 3O (g). Strategy The equation in part (a) depicts two reactants and one product. The equation in part (b) represents a combination of a compound containing C, H, and O with O to produce CO and H O. The equation in part (c) represents two products being formed from a single reactant. These equations represent (a) a combination reaction, (b) a combustion reaction, and (c) a decomposition reaction. Think About It Make sure that a reaction identified as combination has only one product [as in part (a)], a reaction identified as combustion consumes O and produces CO and H O [as in part (b)], and a reaction identified as a decomposition has only one reactant [as in part (c)]. 8. Combustion Analysis Combustion Analysis The experimental determination of an empirical formula is carried out by combustion analysis. In the combustion of 18.8 g of glucose, 7.6 g of CO and 11.3 g of H O are produced. It is possible to determine the mass of carbon and hydrogen in the original sample as follows: 1 mol CO 1 mol C 1.01 g C mass of C = 7.6 g CO 7.53 g C 44.01 g C 1 mol CO 1 mol C 1 mol H O mol H 1.008 g H mass of H = 11.3 g HO 1.6 g H 18.01 g C 1 mol HO 1 mol H The remaining mass is oxygen: 18.8 g glucose (7.53 g C + 1.6 g H) = 10.0 g O Combustion Analysis It is now possible to calculate the empirical formula. Step 1: Determine the number of moles of each element. 1 mol C moles of C = 7.53 g C 0.67 moles C 1.01 g C 1 mol H moles of H = 1.6 g H 1.5 moles H 1.008 g H 1 mol O moles of O = 10.0 g O 0.66 moles O 16.00 g O Step : Write the empirical formula and divide by the smallest subscript to find the whole number ratio. C 0.67 H 1.5 O 0.66 simplifies to CH O 0.66 0.66 0.66 Combustion Analysis The molecular formula may be determined from the empirical formula if the approximate molecular mass is known. To determine the molecular formula, divide the molar mass by the empirical formula mass. For glucose: Empirical formula: CH O Empirical formula mass: [1.01 g/mol + (1.008 g/mol) + 16.00 g/mol] 30 g/mol Molecular mass: 180 g/mol Molecular mass/empirical mass: 180/30 = 6 Molecular formula = [CH O] x 6 = C 6 H 1 O 6 3
1/7/013 Worked Example 8.4 Combustion of a 5.50-g sample of benzene produces 18.59 g CO and 3.81 g H O. Determine the empirical formula and the molecular formula of benzene, given that its molar mass is approximately 78 g/mol. Strategy Determine the mass of C and H in the 5.50-g sample of benzene. Sum these masses and subtract from the original sample mass to find the mass of O. Convert the mass of each sample to moles, and use the results as subscripts in a chemical formula. Convert the subscripts to whole numbers by dividing each number by the smallest subscript to obtain the empirical formula. To calculate the molecular formula, first divide the molar mass given in the problem by the empirical formula mass. Then, multiply the subscripts in the empirical formula by the resulting number to obtain the subscripts in the molecular formula. Worked Example 8.4 (cont.) We calculate the mass of carbon and the mass of hydrogen in the products (and therefore the original 5.50-g sample) as follows: 1 mol CO mass of C = 18.59 g CO 1 mol C 44.01 g CO 1 mol CO mass of H = 3.81 g H O 1 mol H O 1 mol H 44.01 g H O 1 mol H O 1.01 g C 1 mol C 1.008 g H 1 mol H = 5.073 g C = 0.46 g C The total mass of products is 5.073 g + 0.46 g = 5.499 g. Because the combined masses of C and H account for the entire mass of the original sample (5.499 5.50 g), this compound must not contain O. Worked Example 8.4 (cont.) 8.3 Calculations with Balanced Chemical Equations Converting mass to moles for each element present in the compound, moles of C = 5.073 g C moles of H = 3.81 g H 1 mol C 1.01 g C 1 mol H 1.01 g H = 0.44 mol C = 0.43 mol H gives the formula C 0.444 H 0.43. Converting the subscripts to whole numbers (0.44/0.44 1; 0.43/0.44 1) gives the empirical formula CH. Finally, dividing the approximate molar mass (78 g/mol) by the empirical molar mass (1.01 g/mol + 1.008 g/mol = 13.0 g/mol) gives 78/13.0 6. Then, multiplying both subscripts in the empirical formula by 6 gives the molecular formula C 6 H 6. Think About It Use the molecular formula to determine the molar mass and make sure that the result agrees with the molar mass given in the problem. For C 6 H 6, the molar mass is 6(1.01 g/mol) + 6(1.008 g/mol) = 78.11 g/mol, which agrees with the 78 g/mol given in the problem statement. Balanced chemical equations are used to predict how much product will form from a given amount of reactant. moles of CO combine with 1 mole of O to produce moles of CO. moles of CO is stoichiometrically equivalent to moles of CO. Calculations with Balanced Chemical Equations Consider the complete reaction of 3.8 moles of CO to form CO. Calculate the number of moles of CO produced. Calculations with Balanced Chemical Equations Consider the complete reaction of 3.8 moles of CO to form CO. Calculate the number of moles of O needed. mol CO moles CO produced = 3.8 mol CO = 3.8 mol CO mol CO 1 mol O moles O needed = 3.8 mol CO = 1.91 mol O mol CO 4
1/7/013 Worked Example 8.5 Worked Example 8.5 (cont.) Urea [(NH ) 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 5.5 moles of ammonia. (b) Determine the stoichiometric amount of carbon dioxide required to react with 5.5 moles of ammonia. Strategy Use the balanced chemical equation to determine the correct stoichiometric conversion factors, and then multiply by the number of moles of ammonia given. (a) moles (NH ) CO produced = 5.5 mol NH 3 (b) moles CO produced = 5.5 mol NH 3 1 mol (NH ) CO mol NH 3 =.63 mol (NH ) CO 1 mol CO mol NH 3 =.63 mol CO 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 (.63) should be smaller than the number of moles given in the problem (5.5). 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 Dinitrogen monoxide (N 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 Δ NH 4 NO 3 (s) N O(g) + H 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. 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. Worked Example 8.6 (cont.) (a) 10.0 g N O 0.7 mol N O Think About It Use 80.05 the law g NH of 4 conservation NO 3 of mass to check your 0.7 answers. mol NH Make 4 NOsure 3 that 1 mol the combined NH 4 NO 3 mass = 18. of both g NHproducts 4 NO 3 is equal to the mass of reactant you determined in part (a). In this case Thus, 18. (rounded g of ammonium to the appropriate nitrate number must be of heated significant in order figures), to produce 10.0 10.0 g + g of nitrous 8.18 oxide. g = 18. g. Remember that small differences may arise as the result of rounding. (b) Starting with the number of moles of nitrous oxide determined in the first step of (a), mol H 0.7 mol N O O = 0.454 mol H 1 mol N O O 0.454 mol H O 1 mol N O 44.0 g N O 1 mol NH 4 NO 3 1 mol N O 18.0 g H O 1 mol H O = 0.7 mol N O = 0.7 mol NH 4 NO 3 = 8.18 g H O 8.4 Limiting Reactants Limiting Reactants 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) + H (g) CH 3 OH(l) Consider the reaction between 5 moles of CO and 8 moles of H to produce methanol. CO(g) + H (g) CH 3 OH(l) How many moles of H are necessary in order for all the CO to react? mol H moles of H = 5 mol CO = 10 mol H 1 mol CO How many moles of CO are necessary in order for all of the H to react? 1 mol CO moles of CO = 8 mol H = 4 mol CO mol H 10 moles of H required; 8 moles of H available; limiting reactant. 4 moles of CO required; 5 moles of CO available; excess reactant. 5
1/7/013 Worked Example 8.7 Worked Example 8.7 (cont.) 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 (g) + 3H O(l) + Na 3 C 6 H 5 O 7 (aq) The formation of CO causes the trademark fizzing when the tablets are dropped into a glass of water. An Alka-Seltzer tablet contains 1.700 g of sodium bicarbonate and 1.000 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 forms. 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 produced. Finally, use the appropriate molar masses to convert moles of excess reactant and moles of CO to grams. 1 mol NaHCO 1.700 g NaHCO 3 3 = 0.004 mol NaHCO 84.01 g NaHCO 3 3 1.000 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 0.004 mol sodium bicarbonate. 0.004 mol NaHCO 3 1 mol H 3 C 6 H 5 O 7 19.1 g H 3 C 6 H 5 O 7 = 0.00505 mol H 3 C 6 H 5 O 7 1 mol H 3 C 6 H 5 O 7 3 mol NaHCO 3 = 0.006745 mol H 3 C 6 H 5 O 7 The amount of H 3 C 6 H 5 O 7 required to react with 0.004 mol of NaHCO 3 is more than a tablet contains. Therefore, citric acid is the limiting reactant and sodium bicarbonate is the excess reactant. Worked Example 8.7 (cont.) (b) To determine the mass of excess reactant (NaHCO 3 ) left over, first calculate the amount of NaHCO 3 that will react: 0.00505 mol H 3 C 6 H 5 O 7 3 mol NaHCO 3 1 mol H 3 C 6 H 5 O 7 = 0.0156 mol NaHCO 3 Thus, 0.0156 mole of NaHCO 3 will be consumed, leaving 0.0046 mole unreacted. Convert the unreacted amount to grams as follows: 0.0046 mol NaHCO 3 84.01 g NaHCO 3 1 mol NaHCO 3 = 0.388 g NaHCO 3 Worked Example 8.7 (cont.) (c) To determine the mass of CO produced, first calculate the moles of CO produced from the number of moles of limiting reactant (H 3 C 6 H 5 O 7 ) consumed. 3 mol CO 0.00505 mol H 3 C 6 H 5 O 7 Think About It In a problem 1 such mol as H 3 this, C 6 H 5 it Ois 7 a = good 0.0156 idea mol to check CO your work by calculating the amounts of the other products in the reaction. Convert According this amount to the to law grams of conservation as follows: of mass, the combined starting mass of the two reactants (1.700 44.01 g g + CO 1.000 0.0156 mol CO g =.700 g) should equal the sum of the masses of products and = 0.6874 g CO 1 mol leftover CO excess reactant. In this case, the masses of H O and Na 3 C 6 H 5 O 7 produced are 0.815 g and 1.343 g, To summarize respectively. the The results: mass (a) of CO citric acid is the limiting reactant, (b) 0.388 g produced is 0.6874 g [from part (c)] and sodium the amount bicarbonate of excess remains NaHCO unreacted, and (c) 0.6874 g carbon dioxide is 3 is 0.388 g [from part (b)]. The total, produced. 0.815 g + 1.343 g + 0.6874 g + 0.388 g, is.700 g, identical to the total mass of reactants. 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. actual yield % yield = 100% theoretical yield Worked Example 8.8 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: C 7 H 6 O 3 salicylic acid + C 4 H 6 O 3 C 9 H 8 O 4 + acetic anhydride acetylsalicylic acid HC H 3 O acetic acid In a certain aspirin synthesis, 104.8 g of salicylic acid and 110.9 g of acetic anhydride are combined. Calculate the percent yield if 105.6 g of aspirin are produced. 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. 6
1/7/013 Worked Example 8.8 (cont.) 104.8 g C 7 H 6 O 3 110.9 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 0.7588 mol. We convert this to grams using the molar mass of aspirin: 0.7588 mol C 9 H 8 O 4 1 mol C 7 H 6 O 3 138.1 g C 7 H 6 O 3 = 0.7588 mol C 7 H 6 O 3 1 mol C 4 H 6 O 3 10.09 g C 4 H 6 O 3 = 1.086 mol C 4 H 6 O 3 180.15 g C 9 H 8 O 4 1 mol C 9 H 8 O 4 = 136.7 g C 9 H 8 O 4 Worked Example 8.8 (cont.) Thus, the theoretical yield is 136.7 g. If the actual yield is 105.6, the percent yield is % yield = 105.6 g 136.7 g 100% = 77.5% Think About It Make sure you have used the proper molar masses and remember that percent yield can never exceed 100 percent. 8.5 Periodic Trends in Chemical Properties of the Main Group Elements General Trends in Reactivity Ionization Energy and Electron Affinity enable us to understand types of reactions that elements undergo and the types of compounds formed. Hydrogen (1s 1 ) Grouped by itself Forms a cation with a +1 charge (H + ) Forms an anion with a -1 charge (H - ) Hydrides react with water to produce hydrogen gas and a base. CaH (s) + H O(l) Ca(OH) (aq) + H (g) Reactions of the Active Metals Group 1A Elements (ns 1, n ) Low IE Never found in nature in pure elemental state React with oxygen to form metal oxides sodium potassium cesium Reactions of the Active Metals Group A Elements (ns, n ) Less reactive than 1A Some react with H O to produce H Some react with acid to produce H Alkali metals reacting with water Barium reacting with water 7
1/7/013 Group 3A elements (ns np 1, n ) Metalloid (B) and metals (all others) Al forms Al O 3 with oxygen Al forms +3 ions in acid Others form +1 and +3 Group 4A elements (ns np, n ) Nonmetal (C); metalloids (Si, Ge) and others metals Form + and +4 oxidation states Sn, Pb react with acid to produce H Finely-divided aluminum sprinkled into a flame to form Al O 3 Group 5A elements (ns np 3, n ) Nonmetal (N, P), metalloid (As, Sb), and metal (Bi) Nitrogen, N, forms variety of oxides Phosphorus, P 4 As, Sb, Bi (crystalline) Group 6A elements (ns np 4, n ) Nonmetals (O, S, Se) Oxygen, O Sulfur, S 8 Selenium, Se 8 Metalloids (Te, Po) Te, Po (crystalline) HNO 3 and H 3 PO 4 important industrially SO, SO 3, H S, H SO 4 Forest damaged by acid rain Nonmetal oxides added to water produce an acid Group 7A elements (ns np 5, n ) All diatomic Do not exist in elemental form in nature Form ionic salts Form molecular compounds with each other Group 8A elements (ns np 6, n ) All monatomic Filled valence shells Considered inert until 1963 when Xe and Kr were used to form compounds No major commercial use React with hydrogen to form hydrogen halides 8
1/7/013 Comparison of Group 1A and Group 1B Elements 8 Chapter Summary: Key Points Have single valence electron Properties differ Group 1B much less reactive than 1A High IE of 1B - incomplete shielding of nucleus by inner d - outer s electron of 1B strongly attracted to nucleus 1B metals often found elemental in nature (coinage metals) Chemical Equations Interpreting and Writing Chemical Equations Patterns of Chemical Reactivity Combustion Analysis Determination of Empirical Formula Calculations with Balanced Chemical Equations Moles of Reactants and Products Mass of Reactants and Products Limiting Reactants Determining the Limiting Reactant Reaction Yield Periodic Trends in Reactivity of the Main Group Elements General Trends in Reactivity Reactions of the Active Metals Reactions of the Other Main Group Elements Comparison of Group 1A and Group 1B Elements Group Quiz #14 Group Quiz #15 Balance the following equations: Fe (SO 4 ) 3 + K 3 PO 4 K SO 4 + FePO 4 Hg(NO 3 ) + KI HgI + KNO 3 Li O(s) + H O LiOH HBr + Ba(OH) BaBr + H O K PtCl 4 + NH 3 Pt(NH 3 ) Cl + KCl How many atoms are in 1.987 g of iron? How many H atoms are in a 3.415 gram sample of H O? Group Quiz #16 Complete the equation below (with phases) and balance it. K S(aq) + AgNO 3 (g) If you combine 10.1 ml of 0.15 M K S with 1.009 g AgNO 3, how much solid product can be formed? If 0.744 g of product were actually formed, what is the percent yield? 9