ExtraChem2-20-2014.pdf

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Department
Chemistry
Course
CHE 2A
Professor
A N D R E A S T O U P A D A K I S
Semester
Winter

Description
10.  When  asking  for  the  %  by  mass  of  a  certain  compound,  you  need  to Given:  Na2CO3  and  NAHCO3 mass=45.0g  of  mixture mixture  reacted  with  excess  HNO3  solution  and  the  produced  gas  occupies  10L(V)  85  C(T) 1.32  atm  (P) A)  xg  of  NA2CO3  Calculate  molar  mass=  106.0g/mol xg  NaHCO3  M.M.=  84.01g/mol B)x+y=45g C)T=85+273=358K D)Now  find  the  moles  produced  total  from  ideal  gas  law  equation=  PV=nRT PV/RT=n  .449  mol  CO2 Na2CO3+2HNO3  →  2NaNO3+CO2+H2O (x/106.0mol)  →    (x/106.0mol) NaHCO3  +  HNO3  →  NaNO3+CO2+H2O (y/84.01  mol)  →    (y/84.01  mol) (x/106)+(y/84)  →  .449  because  x=45^-­y y=9.91/45.0  x100=  22% 11.  when  looking  for  the  molar  mass  of  the  base  in  a  titration given:  mass=1.58g  of  diprotic  base dissolved  in  water resulting  solution  is  titrated  with  .0100M  HCl  solution V=60.5mL  acid  solution  required  to  neutralize  the  base A)  (60.5mL  HCl)(.0100M  HCl)  =  .605  mmol  HCl find  moles  by  multiplying  the  volume  of  the  acid  you  titrate  with  and  the  potency  of  the  HCl solution B)  2HCl+M(OH)2  →  MCl2+2H2O next  write  balanced  equation  of  the  titration In  a  reaction  between  an  acid  and  base  yields  a  salt  and  water C)  Now  find  the  mol  of  the  metal  hydroxide because  of  stoichiometric  ratio (.605  mmol  HCl)  x  (1mmol  of  M(OH)2/  2  mmol  HCl)  =  .303  mmol  M(OH)2 D)  Now  that  you  have  the  mmol  of  the  Base  reactant  you  can  find  the  salt  you  can  find  molar mass-­ Molarity n=m/M.M. M.M.=m/n  =  1.58g/3.03x10  =  5.21  x  10  g/mol 3 The  base  could  be  a  large  organic  molecule. 1)  Which  one  of  these  rxns  yield  a  precipitate.. Need  to  know  what  com
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