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A solution of 0.393 m ba(oh)2 required 32.24 ml of a 0.227 m solution of hn03 calculate the original volume of the ba(oh)2 soluion

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kenmyna
The  original  volume    of  the  Ba(OH)2  solution  is calculated  as  follows


find the  moles   of HNO3  used

Ba(OH)2  +  2HNO3  =Ba(NO3)2  +2H2O

calculate  the  moles   of   HNO3  used  =molarity  x  volume
= 32.24  x0.227=  7.32   moles

by  use of  mole  ratio  between  Ba(OH)2 to HNO3  which  is   1:2 the moles  of  Ba(OH)2  is  therefore  =  7.32  x1/2 =3.66   moles

volume of Ba(OH)2   is  therefore = (3.66  /0.393= 9.3 ML  of Ba(OH)2
carlyfleetwood67

Answer:(3.66  /0.393= 9.3 ML  of Ba(OH)2

Explanation: The  original  volume    of  the  Ba(OH)2  solution  is calculated  as  follows find the  moles   of HNO3  used

Ba(OH)2  +  2HNO3  =Ba(NO3)2  +2H2O

calculate  the  moles   of   HNO3  used  =molarity  x  volume

= 32.24  x0.227=  7.32   moles

by  use of  mole  ratio  between  Ba(OH)2 to HNO3  which  is   1:2 the moles  of  Ba(OH)2  is, therefore,  =  7.32  x1/2 =3.66   moles

volume of Ba(OH)2   is  therefore = (3.66  /0.393= 9.3 ML  of Ba(OH)2

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