Chemistry question *help*

• SIMpLiFy

  23 Jan 08, 20:40

  In an experiment,excess zinc powder reacts with 50cm cube of 0.1dm cube/mol sulphuric acid in an experiment.The products formed in the reaction are zinc sulphate and hydrogen gas.
  
  (a)Write a balanced equation for the reaction.
  (b)Determine the volume of hydrogen gas produced in the reaction.
  
  Thanks.
• SBS3625D

  23 Jan 08, 21:01

  WTF.
  
  You might as well give me a table, a calculator and your exam paper?
  
  Do it yourself please....there's only so much we can help you.
• wishboy

  23 Jan 08, 21:42

  Originally posted by SIMpLiFy:

  In an experiment,excess zinc powder reacts with 50cm cube of 0.1dm cube/mol sulphuric acid in an experiment.The products formed in the reaction are zinc sulphate and hydrogen gas.
  
  (a)Write a balanced equation for the reaction.
  (b)Determine the volume of hydrogen gas produced in the reaction.
  
  Thanks.

  a)
  Zn (s) + H2SO4 (aq) ----> ZnSO4 (aq) + H2 (g)
  
  b)
  No. of mol H2SO4
  = 0.05/0.1
  = 0.5 mol
  
  Vol. of H2
  = 0.5*24
  = 12 dm³
  
  i not sure whether my answers correct anot
  anyway try to understand how to do
  anything dunno ask ur teacher/classmate/anyone who noes how to do
• SIMpLiFy

  24 Jan 08, 21:49

  ok..sry , i just noticed my teacher havn't taught this topic yet , nowonder it looks so unfamiliar..well thx anyway =D.
• gnab

  24 Jan 08, 22:05

  6Zn (s) + 2HSO4 (aq) -> 2Zn3SO4 (s) + H2 ( g)
  
  no of mol of HSO4 used = (50/1000)/.1 = 0.5mol
  no of mol of H2 produced = no of mol of HSO4 = 0.5mol
  
  hope my answer is useful..
  
  
• wishboy

  24 Jan 08, 22:26

  Originally posted by gnab:

  6Zn (s) + 2HSO4 (aq) -> 2Zn3SO4 (s) + H2 ( g)
  
  no of mol of HSO4 used = (50/1000)/.1 = 0.5mol
  no of mol of H2 produced = no of mol of HSO4 = 0.5mol
  
  hope my answer is useful..
  
  

  er
  how come in ur equation sulphuric acid is HSO4? not H2SO4 meh?
• BaoBei92

  5 Mar 08, 18:34

  Originally posted by wishboy:

  er
  how come in ur equation sulphuric acid is HSO4? not H2SO4 meh?

  6Zn (s) + 2HSO4 (aq) -> 2Zn3SO4 (s) + H2 ( g)

  he used 2HSO4. probably typo.

• Darkness_hacker99

  6 Mar 08, 06:19

  This topic is about Limiting Reactant and Chemical Calculation. Limiting reacting as in 'not in excess'.
  
  First must find out which is the limiting reactant,  before you can do the calculation. Since zinc power is in excess, then sulpuhruic will be the limiting reactant.

  
  Next is to find the no. of mol of sulphuric acid. With that, you can then calculate volume of hydrogen gas produced.

• Darkness_hacker99

  6 Mar 08, 06:22

  It's H₂SO₄ at O Level.

• Darkness_hacker99

  6 Mar 08, 16:07

  Originally posted by wishboy:

  a)
  Zn (s) + H2SO4 (aq) ----> ZnSO4 (aq) + H2 (g)
  
  b)
  No. of mol H2SO4
  = 0.05/0.1
  = 0.5 mol
  
  Vol. of H2
  = 0.5*24
  = 12 dm³
  
  i not sure whether my answers correct anot
  anyway try to understand how to do
  anything dunno ask ur teacher/classmate/anyone who noes how to do

  Your answer correct.

• weewee

  6 Mar 08, 16:15

  Originally posted by wishboy:

  a)
  Zn (s) + H2SO4 (aq) ----> ZnSO4 (aq) + H2 (g)
  
  b)
  No. of mol H2SO4
  = 0.05/0.1
  = 0.5 mol
  
  Vol. of H2
  = 0.5*24
  = 12 dm³
  
  i not sure whether my answers correct anot
  anyway try to understand how to do
  anything dunno ask ur teacher/classmate/anyone who noes how to do

  Why is it *24?

  Holy i have forgotten my chemistry.

• weewee

  6 Mar 08, 16:20

  now i remember the formula 1 mole = 24dm3 but forgot why.

• Darkness_hacker99

  6 Mar 08, 16:21

  Molar Gas Volume = 24dm³
  
  It's the amount of volume occupied by 1 mole of gas molecule @ r.t.p

• eagle

  6 Mar 08, 16:22

  Originally posted by weewee:

  now i remember the formula 1 mole = 24dm3 but forgot why.

  0.5 mole of H2SO4 still consists of 0.5 mole of H2.

  So volume of H2 produced from 0.5 mol of H2SO4 is still 0.5 mol of H2 (because it is by number of molecules).

  Since 1 mole = 24dm3, 0.5 mol = 12 dm3 of H2