Chem Stoichiometry Practice Solutions 3

 To determine the molar mass of an unknown metal, M, a student reacts iodine with an excess of the metal to form the water-soluble compound MI , as represented by the equation above. The reaction proceeds until all of the I is consumed. The MI (aq) solution is quantitatively collected and heated to remove the water, and the product is dried and weighed to constant mass. The experimental steps are represented below, followed by a data table. Data for Unknown Metal Lab Mass of beaker 125.457 g Mass of beaker + metal M 126.549 g Mass of beaker + metal M + I 127.570 g Mass of MI , first weighing 1.284 g Mass of MI , second weighing 1.284 g Question (a) Given that the metal M is in excess, calculate the number of moles of I that reacted. 127.570  126.549 = 1.021 g I 1.021 g I × 1 mol I 2 = 0.004023 mol I 253.80 g I 2 1 point is earned for the number of moles. (b) Calculate the molar mass of the unknown metal M. Number of moles of I = number of moles of M 1.284 g MI  1.021 g I = 0.263 g M Molar mass of M = 0.263 g M = 65.4 g/mol 0.004023 mol M 1 point is earned for the number of grams of M. 1 point is earned for the molar mass. The student hypothesizes that the compound formed in the synthesis reaction is ionic. (c) Propose an experimental test the student could perform that could be used to support the hypothesis. Explain how the results of the test would support the hypothesis if the substance was ionic. The student could dissolve the compound in water or melt the compound and see if the solution/melt conducts electricity. If the solution/melt conducts electricity, mobile ions capable of carrying charge must be present, thus the compound is likely to be ionic. OR The student could heat the compound until it melts or boils. If the melting/boiling point is very high, then the compound is likely to be ionic. 1 point is earned for an appropriate test. 1 point is earned for explaining how the results would support the hypothesis. The student hypothesizes that Br will react with metal M more vigorously than I did because Br is a liquid at room temperature. Question (d) Explain why I is a solid at room temperature whereas Br is a liquid. Your explanation should clearly reference the types and relative strengths of the intermolecular forces present in each substance. Both Br and I molecules are nonpolar molecules, therefore the only possible intermolecular forces are London dispersion forces. The London dispersion forces are stronger in I because it is larger in size with more electrons and/or a more polarizable electron cloud. The stronger London dispersion forces in I result in a higher melting point,which makes I2 a solid at room temperature. 1 point is earned for identifying the forces in each substance as London dispersion forces. 1 point is earned for explaining why the forces are stronger in I than in Br . While cleaning up after the experiment, the student wishes to dispose of the unused solid I in a responsible manner. The student decides to convert the solid I to (aq) anion. The student has access to three solutions, H O (aq), Na S O (aq), and Na S O (aq), and the standard reduction table shown below. Half-reaction SO (aq) + 2 e  2 S O E° (V) (aq) 0.08 I (s) + 2 e  2 I(aq) 0.54 O (g) + 2 H (aq) + 2 e  H O (aq) 0.68 (e) Which solution should the student add to I (s) to reduce it to I (aq)? Circle your answer below. Justify your answer and include a calculation of E° for the overall reaction. H O (aq) Na S O (aq) [Na S O (aq) should be circled.] Na S O (aq) 1 point is earned for the correct choice. (aq) The reaction between S O and I (s) will be thermodynamically favorable because or the reaction is positive ( = 0.54  0.08 = +0.46 V), from which it follows that   is negative because   = nF . 1 point is earned for a correct justification. (f) Write the balanced net-ionic equation for the reaction between I and the solution you selected in part (e). I +2S O   2 I + S O  1 point is earned for the correct equation.