practice calculations

Practice Calculations:                                                                                                                     Write out the known and unknown data first. Show all working and record answers to 3 significant figures.   1.    An acid – base titration yielded the following information. 1.25.0 mL of a 0.250 molL-1solution of potassium hydroxide required 15.4 mL of sulfuric acid to be fully neutralised. Using the equation below calculate the concentration of the sulfuric acid  in molL-1 2KOH + H2SO4  ® K2SO4  + 2H2O   2.    An acid – base titration yielded the following information.       13.7ml of 0.250molL-1 HCl was required to fully neutralise 25ml of Ba(OH)2. Using the                  equation below calculate the concentration of the Ba (OH)­2 in molL-1.    Ba(OH)2 + 2HCl     ®       BaCl2 +   2H2O   3.    Aspirin, C9H8O4 , is made by reaction of:               Salicyclic Acid + Ethanoic anhydride à       Aspirin + Ethanoic acid   C7­H6O3  +  C4­H6O3     à        C9H8O4  +  C2­H4O2   How many grams of ethanoic anhydride C4­H6O3 are needed to make one aspirin            tablet, which contains 0.33 g of aspirin?     4.    When 4.78g of an oxide of lead are heated in a stream of hydrogen, 4.14g of lead are formed.                                PbxOy      +    H2    à      Pb   + H2O               Determine the empirical formula of the lead oxide, PbxOy.             You need to work out the mass of the lead and oxygen so that you can determine the       empirical formal of PbxOy                                 Help             The mass of lead, Pb in PbxOy = 4.14g             The mass of lead oxide, PbxOy = 4.78g             So mass of oxygen in PbxOy must be  4.78g - 4.14g = 0.64 g                    n(Ba(OH)2) = 3.425 x 10‑3 mol = 1.7125 x 10-3 mol c(Ba(OH)2) = = 0.0685molL-1   Q3) M(C9H8O4) = 180 g mol-1 (i)     n(C9H8O4) = m/M  = 0.33 g / 180 g mol-1  = 1.83 x 10-3 mol (ii)    n(C4H6O3) = n(C9H8O4) = 1:1 mole ratio so         n(C4H6O3) 1.83 x 10-3 mol (iii)       m(C4H6O3) = 1.83 x 10-3 mol x 102 g mol-1 = 0.187 g     Q4) Mass of lead oxide is 4.78g Mass of lead is 4.14g So mass of oxygen that is lost by the lead is worked out from the mass of the PbO – mass of Pb The mass of oxygen is  4.78g - 4.14g = 0.64 g  We don’t need the mass of H2 or H2O   i.e.    PbO       +    H2    à      Pb   + H2O (4.78g + 0.64 g)   + H2  à    4.14g   +   H2O   i)      n(Pb) = 4.14/207= 0.020 mol (ii)    m(O) = 4.78 - 4.14 = 0.64 g         (iii)   n(O) = 0.64/16= 0.04 mol (iv)       Hence Pb : O  = 1 : 2 and empirical formula is PbO2.

Answers:Q1)n(KOH) = 0.250 x 25/1000 = 0.00625 moln(H2SO4) = n(KOH)n(H2 SO4) =  x 0.00625 mol = 0.003125 molc(H2SO4) =  = 0.203 molL-1  Q2)n(HCl) = 0.250 x 0.0137 = 3.425 x 10‑3 moln(Ba(OH)2) = n(HCl)The mass of oxygen is  4.78g - 4.14g = 0.64 g We don’t need the mass of H2 or H2O i.e.    PbO       +    H2    à      Pb   + H2O(4.78g + 0.64 g)   + H2  à    4.14g   +   H2O i)      n(Pb) = 4.14/207= 0.020 mol(ii)    m(O) = 4.78 - 4.14 = 0.64 g        (iii)   n(O) = 0.64/16= 0.04 mol(iv)       Hence Pb : O  = 1 : 2 and empirical formula is PbO2.

practice calculations