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Calculate the molarity of the sulphuric acid.

  1. May 23, 2007 #1
    Hello people...could u help me pls...coz i am rly stuck in these thing like calculations...i know all right answers but i am not too sure how to get them...please if somebody of us has a few minutes could u help me with that?
    Thank you very much.

    1.,
    When 0.25 g of sodium metal was added to 200 cm3 ( an excess ) of water , the following reaction occured.

    Na(s) + H20 (l) --> NaOH ( aq ) + 1/2 H2 (g )

    (i) calculate the number of moles of sodium taking part in the reaction.

    Na = 23 g mol-1
    0.25g / 23 g mol-1 = 0.011 mole

    (ii) calculate the molarity of the sodium hydroxide solution which was
    formed.

    (iii) calculate the volume of hydrogen gas produced at r.t.p. (298 K and
    100 kPa ). Assume that hydrogen is insoluble in water under these
    conditions.

    In another experiment 25.0 cm3 of 0.183 M sodium hydroxide were neutralised by 13.7 cm3 of sulphuric acid according to the following equation.

    2NaOH (aq) + H2SO4 (aq) --> Na2SO4 (aq) + 2H2O ( l )

    Calculate the molarity of the sulphuric acid.
     
    Last edited: May 23, 2007
  2. jcsd
  3. May 23, 2007 #2

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    Again, please show your work.
     
  4. May 23, 2007 #3
    oki doki...sorry i didn t know about this rule.
     
  5. May 23, 2007 #4
    calculation...could somebody help me please???

    Hello people...could u help me pls...coz i am rly stuck in these thing like calculations...i know all right answers but i am not too sure how to get them...please if somebody of us has a few minutes could u help me with that?
    Thank you very much.

    1.,
    When 0.25 g of sodium metal was added to 200 cm3 ( an excess ) of water , the following reaction occured.

    Na(s) + H20 (l) --> NaOH ( aq ) + 1/2 H2 (g )

    (i) calculate the number of moles of sodium taking part in the reaction.

    Na = 23 g mol-1
    0.25g / 23 g mol-1 = 0.011 mole

    (ii) calculate the molarity of the sodium hydroxide solution which was
    formed.

    (iii) calculate the volume of hydrogen gas produced at r.t.p. (298 K and
    100 kPa ). Assume that hydrogen is insoluble in water under these
    conditions.

    In another experiment 25.0 cm3 of 0.183 M sodium hydroxide were neutralised by 13.7 cm3 of sulphuric acid according to the following equation.

    2NaOH (aq) + H2SO4 (aq) --> Na2SO4 (aq) + 2H2O ( l )

    Calculate the molarity of the sulphuric acid.
     
  6. May 23, 2007 #5

    chemisttree

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    Science Advisor
    Homework Helper
    Gold Member

    (ii) How many moles, relative to sodium, of sodium hydroxide are produced in the reaction?

    What is the definition of molarity?

    (iii) How many moles of hydrogen, relative to sodium, are produced? What equation might you use to convert this number of moles into a volume?

    (other experiment) How many moles of NaOH are used in this neutralization? How does this number of moles of NaOH relate to the number of moles of H2SO4?
     
  7. May 31, 2007 #6

    Borek

    User Avatar

    Staff: Mentor

    Janka, please read about balanced equation meaning. Especially 2nd paragraph on the page. Once you will master this information, stoichiometry should become obvious :)
     
  8. Jun 10, 2007 #7
    hey janka.....
    hhmm... the first one is good...
    molarity means how many moles of a substance is present in a definite volume. as you know it, we often use mol/dm3

    here you have 200cm3 of water. the NaOH will be present in there. now, you only need to know how much of NaOH is in the 200cm3 water.

    from the first equation, 1 mol of Na gives 1 mol of NaOH. but you are using 0.011 mol Na, so you will get 0.011 mol NaOH (by proportion).

    then, it should be true that 200cm3 of solution (water + NaOH) contains 0.011 mol NaOH. therefore, in 1000cm3 solution, you will get (0.011*1000)/200 = 0.055 mol NaOH.

    you have just calculated the molarity: 0.055 mol/dm3

    using the same proportion method: from 1 mol Na you get 0.5 mol H2. But you are using 0.011 mol Na, so you'll get 0.0055 mol H2.

    at stp, 1 mol of a gas occupies 22.5dm3
    0.0055 mol H2 will occupy, 22.5*0.0055 = 123.75 cm3



    from the other exp:

    you can calculate the amt of NaOH that reaced.
    1000cm3 contains 0.183 mol NaOH
    25 cm3 will contain 0.005 (approx.) mol NaOH

    according to the eqn, 2 mol NaOH reacts with 1 mol H2SO4
    0.005 (using the true value) mol NaOH will react with 0.002 mol H2SO4

    therefore, 13.7 cm3 of H2SO4 conatins approx. 0.002 mol
    using the exact value, 1000cm3 H2SO4 will contain 0.17 mol = approx. 0.2 mol H2SO4

    i'm getting the feeling tat i already replied to this question.... nyway... it won;t kill anyone if i do it again...lool
     
  9. Jun 10, 2007 #8
    i may b sooo stupid sometimes lool.... i've already answered this question in the other thread....

    eerrmmm..... in this one i've made an error though....
    it took stp instead of rtp....
    so the volume of H2 would be 132 cm3 instead of 123.75 cm3
     
  10. May 31, 2008 #9
    Hello Kushal

    Hi Kushal,
    thanks a lot for your help, i was doing this calculations again this year, it really helped me, Thank you !!!
     
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