Practice Problems– Chem II Classic Unit I: Precipitation Reactions & Stoichiometry Name: KEY Period: 1. You dissolve ordinary table salt in water then add to it a solution of aqueous silver nitrate. (Silver nitrate is formed by dissolving solid silver in nitric acid; this feat was first accomplished by Albertus Magnus in the 13th century. Silver nitrate will strongly stain your skin, but is effective against warts.) a. A precipitate forms. Referencing the solubility rules, tell me what this precipitate is made of. Silver chloride is insoluble, so it forms the precipitate. In general, salts with halide anions are soluble, but there is an exception if the cation is silver (Ag+). b. Write out the balanced precipitation reaction in full, including state symbols. Then write out the net ionic reaction (eliminating any spectator ions from consideration). AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq) Ag+ (aq) + Cl− (aq) → AgCl (s) c. Which of the following best describes the reaction above? Circle the right answer. combination decomposition single-­‐replacement polymerization hydrolysis double-­‐replacement acid-­‐base combustion d. Qualitatively describe the entropy and enthalpy changes the system undergoes in this reaction. The entropy decreases because a solid is formed (solids are more highly organized). Since the reaction is spontaneous, this means enthalpy must decrease as well (the reaction is exothermic). e. Let’s look at this quantitatively: you had 5.00 grams of table salt that you added to 100. mL of water to make your first solution. Your silver nitrate solution was 0.60 M and you added 50.0 mL of it to your table salt solution. Determine: (i) which reactant is limiting; (ii) how much of your precipitate should form (in grams); and (iii) how many excess ions will remain in solution (in moles). 5.00 𝑔 𝑁𝑎𝐶𝑙×
50.0 𝑚𝐿 ×
! !"# !"#$
!".!! !
1 𝐿
1000 𝑚𝐿
= 0.0856 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙 0.6 𝑚𝑜𝑙 𝐴𝑔𝑁𝑂!
= 0.0300 𝑚𝑜𝑙 𝐴𝑔𝑁𝑂! 1 𝐿
From the balanced equation, we see they mix in equal ratios, so silver nitrate will be the limiting reactant. We will form 0.0300 moles of precipitate (again, the product is in an equal ratio to each of the reactants). The molecular weight of AgCl is 143.32 g/mol, so this amounts to 4.30 g of silver chloride. We are left with 0.0300 moles of nitrate ions in solution. Since we used 0.0300 moles of Cl to form our precipitate, we are left with 0.0856 – 0.0300 = 0.0556 moles of it in solution. We still have all 0.0856 mol of Na in our solution. Charges balance, because we have 0.0856 mol of Na+ and 0.0556 + 0.0300 moles of Cl-­‐ and NO3-­‐, respectively, left in solution. Practice Problems– Chem II Classic Unit I: Precipitation Reactions & Stoichiometry f.
Name: KEY Period: Draw a sub-­‐microscopic depiction of the resulting mixture after mixing the reactants described in part (e). Considerations: a simple ionic lattice of Ag and Cl should be drawn as settled to the bottom of the container; the aqueous solution itself should contain no silver, but should include hydrated sodium, chloride, and nitrate ions in roughly 3: 2 : 1 ratios, based on their mole ratios above. Free water molecules should also be included in the sketch. 2. For additional practice, repeat Problem 1 but instead of table salt use beryllium iodide, and instead of silver nitrate use copper(I) sulfate. Copper(I) iodide is insoluble, because (again) it is an exception to the halide rule. The balanced and net ionic equations look like this: BeI2 (aq) + Cu2SO4 (aq) → 2Cul (s) + BeSO4 (aq) Cu+ (aq) + I− (aq) → CuI (s) This is, again, a double-­‐replacement reaction, and the entropy & enthalpy considerations are the same as in problem 1. Assuming we again have 5.00 g of BeI2 and 50.0 mL of 0.60 M copper(I) sulfate solution, we have 0.0190 moles of BeI2 and 0.0300 moles of Cu2SO4. So the limiting reactant this time is BeI2. We will create twice as many moles of CuI precipitate as we had BeI2 moles thanks to the mole ratio in the balanced equation, meaning we’ll have 0.0190 moles x 2 = 0.038 moles; using the molecular mass of CuI, we arrive at 7.24 g. We’ll have 0.0190 moles of Be2+ left over in solution. We’ll have 0.0300 – 0.0190 = 0.0110 moles of copper ion pairs, so 0.0220 moles of copper ions remaining in solution. We’ll have 0.0300 moles of the spectator sulfate ion left over. 3. Here are lists of some common cations and anions. Cations: Li+, Ag+, Pb+, NH4+, Cu+, Na+, K+, Al3+ Anions: sulfate, nitrate, chloride, acetate, sulfide, carbonate, phosphate, hydroxide a. Using the solubility rules you were given in class, choose two ionic salts that each are soluble in water, but when combined form a precipitate. Then write out the balanced reaction with state symbols. Ammonium phosphate and lead chloride is one – phosphates tend to be insoluble unless they are attached to ammonium or an alkali metal. There are, of course, many right answers to this question. b. Choose two ionic salts that are each soluble in water, but when combined do NOT form a precipitate. Then write out the balanced reaction with state symbols. Lead acetate and copper nitrate – all nitrates are soluble, and only silver acetates are insoluble. There are, again, many right answers to this question.