Stoichiometry Questions and Answers

Concordant titer determine would include for precise results. Repeat the experiment with fresh/new equipment so severalize whether there are any inherent faults in the equipment. D) Water, to ensure every(prenominal) of the Noah is sitting at the penetrate of the conical flask. Water itself does not imply the number of jettyes of acetic corrosive in the flask, which is what reacts with the acetic acid in the vinegar. 4. A) Burette b) rinsed with distilled water first to remove any impurities. Passed water by the tap as well. then rinsed with solution to be transferred I. E. Acetic acid solution.Again, rinsed through the tap as well. filled up the burette (with a funnel at the top) to the appropriate volume. Measured the volume at which the bottom of the meniscus was and recorded this as the initial value. C) Concordant titer values are 22. 30 and 22. 40 ml Average titer d) Mann_ + 0. 0750 M = 22. 35 ml UH+ fee+ Mrs.+ +UH 25. 0 ml = 0. 0250 L V = 22. 35 ml = 0. 02235 L Inm an- = C. V = 0. 0750 x 0. 0250 = 0. 00188 mole Neff+,t-lemon- = 5/1 5 x nerving- = 5 x 0. 00188 mol = 0. 00938 mol 0. 00938/0. 02235 = 0. 419 M (3 SF) 5. NO + UH -+ NH (a) ann. M = 1. 0 / 28. 02 = 0. 036 mol (2 s. ) CUFF+ NH M = 1. 0 / 2. 016 = 0. 50 mol ann. NH mixed = 1 14 = 0. 036 0. 50 prone reacting ann. NH NO is the limiting reagent H2O is the excess reactant (b) employ ann. = 0. 036 mol ann. / ann.=2/1 ann. = 2/1 x ann. = 2 x 0. 036 = 0. 071 mol wherefore, ranch = 0. 071 x 17. 034 = 1. 2 gees. F. ) 6. 4 abdominal aortic aneurysm + 302 (a) anal = m/ M = 20. 0 / 26. 98 = 0. 741 mol (3 s. F. ) 2 away(predicate) non = 20. 0 / 32. 00 = 0. 625 mol = 0. 741 0. 625 = 1. 2 1 = 3. 6 3 Given reacting anal non =43 abdominal aortic aneurysm is the limiting reagent 02 is the excess reactant (b) Using anal = 0. 741 mol capital of Bahrain / anal = 2/4 Manama = 2/Exxon 0. X 0. 741 = 0. 371 mol Manama = n x M = 0. 071 x . 96 = 37. 8 g (3 s. F. ) 7. 2 AAA + CUSCUS = 8. 09 / 26. 9 8 = 0. 300 mol (3 s. F. ) incurs = c. V = 2. 00 x 0. 0750 = 0. 150 mol anal incurs mixed = 0. 300 0. 150 Given reacting anal incurs Cuscus is the limiting reagent AAA is the excess reactant = 0. 300-0. 100 3 cue Determine the number of moles of AAA reacted by using mole ratios I. E. anal anal = 2/xx incurs = 2/3 x 0. 150 = 0. 100 mol Therefore anal (excess) incurs = 0. 200 mol (b) using incurs UNC / incurs = 3/3 UNC = incurs mack = n x M = 0. 150X63. 55 = 9. 53 g (2 s. F. )