How many liters of h2 would be formed at 600 mm hg and 13 ∘c if 26.5 g of zinc was allowed to react?

The  number  of  liters  of H2  that  would  be  formed  at 600 mmhg and 13  c  if 26.5 g zinc  was  allowed  to react  is  calculated  as  follows

find the  moles of Zn   reacted
that  is 26.5 g /  56.39 g/mol =  0.4699  moles

write  the reaction  Zn  with an  acid  example HCl  to  produce H2

that is  Zn + 2 HCl =  ZnCl2  + H2

by  use of  mole  ratio  between  Zn to H2   the  moles  of H2  = 0.4699  moles

By  use of  Ideal  gas  equation Volume  of H2 =nRT/P

n= 0.499 moles
R= 62.364  L.mmhg/mol.K
T=13 +273=286 k
P= 600 mmhg

= (0.4699  mol x 62.364 L.mmhg/mol.k  x286 k)/600  mm hg  =  13.97 liters of H2

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