1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculation could somebody help me please?

  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.
     
  2. jcsd
  3. May 28, 2007 #2
    Concentration = moles/volume

    Molarity = Concentration

    If you need further help ask away...
     
  4. May 29, 2007 #3
    okay... i'll explain it to you...

    (1) Ar Na: 23 therefore, 0.25 g Na contains (0.25/23) mol = 0.011 mol

    (2) number of moles NaOH formed:

    1 mol Na gives 1 mol NaOH
    therefore, 0.011 mol Na will give 0.011 NaOH

    200 cm3 (approx.) contains 0.011 mol NaOH
    100 cm3 will contain (0.011*100/200) = 0.005 mol (approx.)

    this is the molarity, 0.005 mol/dm3

    (3) NUMBER OF MOLE OF h2 PRODUCED: (0.5*0.011) since 1 mol Na produces 0.5 mol H2 = 0.005 mol
    at r.t.p. 1 mol of gas occupies 24 dm3
    therefore 0.005 mol H2 will occupy (0.005*24) = 0.130 dm3 = 130 cm3


    number of mol NaOH in 25 cm3 = (0.183*25/1000) since 1000 cm3 contains 0.183 mol NaOH = 0.005 mol

    therefore, since 2 mol NaOH reacts with 1 mol H2SO4, the number of mol of H2SO4 which reacted is (0.005/2) = 0.002 mol

    13.7 cm3 contains 0.002 mol H2SO4
    1000cm3 will therefore contain (0.002*1000/13.7) = 0.167 mol H2SO4

    0.167 M
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Calculation could somebody help me please?
  1. Please someone help me (Replies: 3)

Loading...