Respuesta :

kenmyna
The  volume  of  o.378 M  of  Hl  is  calculated  as  follows
find  the  moles  of  Ba(OH)2   used

that  is  moles  =  molarity   x  volume  /1000
 =   21.6  x   0.520/1000= 0.0112  moles

write  the  equation  for  reaction

Ba(oh)2   +2Hl  ------>   Bal2  +  2H2O

fro  the  equation  above  the  reacting  ratio   between  Ba(OH)2  to  Hl  is   1:2   therefore  the  moles  of  Hl  =  2   x0.0112= 0.0224  moles  of  Hl

volume  of  HCl  =  moles  of  Hl / molarity of  Hl  x1000  

that  is   0.0224/0.378x1000=  59.3  ml

superman1987
first, we have to get moles of Ba(OH)2 that reacts with HCl according to this equation:
Ba(OH)2(aq) + 2HCl (aq) → BaCl2(aq) + 2H2O(l)

when moles of Ba(OH)2 = molarity * volume
                                         = 0.52 M * 0.0216 L
                                         = 0.011232 moles
from the balanced equation we can see that the mole ratio between
 HCl : Ba(OH)2 = 2:1
So moles of HCl = 2* moles of Ba(OH)2
                            = 2* 0.011232 moles
                            = 0.0225 moles
now, we can get the volume:
when volume = no of moles of HCl / molarity of HCl
                        = 0.0225 moles / 0.378 M
      ∴volume   = 0.0595 L = 59.5 mL 

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