Turn on thread page Beta

Need help with moles calculation question! watch

Announcements
    • Thread Starter
    Offline

    0
    ReputationRep:
    Heya! I'm a bit confused about a question on my homework and was wondering if anyone can help me out. I've put my working out in red below each question. If someone could see if my answers are correct and help me out with the last one especially, that would be much appreciated! Thanks!


    1) A tank contained 400m3 of waste hydrochloric acid. It was decided to neutralise the acid by adding slaked lime, Ca(OH)2.

    a) The concentration of the acid was first determined by titration of a 25cm3 sample against 0.200mol.dm-3 sodium hydroxide, of which 31.3cm3 were required.

    i) Calculate the concentration of the hydrochloric acid in the sample in mol.dm-3 giving your answer to 3 significant figures.

    mol = 0.0313 x 0.200 = 0.00626 mol of NaOH


    1 NaOH : 1 HCl so 0.00626 mol NaOH : 0.00626 mol HCl

    concentration = 0.00626 / 0.025 = 0.2504 mol dm-3


    ii) Calculate the total number of moles of HCl in the tank. Give your answer to 3 sig.fig.

    400m3 = 40dm3

    mol = 40 x 0.250 = 10 mol of HCl


    b) Calculate the mass in tonnes of slaked lime required to neutralise the acid. Slaked lime reacts with hydrochloric acid according to the following equation: Ca(OH)2 + 2HCl --> CaCl2 + 2H2O

    1Ca(OH)2 : 2HCL

    mol = 10 x 2 = 20 mol Ca(OH)2

    20 = mass / 74.1 = 1482g = 0.00148 tonnes of Ca(OH)2 needed


    c) The slaked lime was manufactured by roasting limestone and then adding water:
    CaCO3 --> CaO + CO2
    CaO + 2H2O --> Ca(OH)2
    Calculate the mass of limestone required to produce 1kg of slaked lime.

    I have no idea how to do this one
    Offline

    21
    ReputationRep:
    You did all the hard ones, the last one is probably the easiest!

    First work out how many moles of Ca(OH)2 you need to make.

    Use Mass = Mr x Mol, where mass = 1kg

    Spoiler:
    Show
    1000 = (40.1 + 34) x mol

    mol = 1000/(40.1+34)

    mol = 13.4953...


    Use this answer to work out the moles of Limestone needed

    Spoiler:
    Show
    Everything relevant uses a 1:1 mole ratio, so there is 13.4953 moles of Limestone required.


    Now using moles of Limestone to work out mass

    Spoiler:
    Show
    mass = mr x mol

    mass = 13.4953... x 100.1
    = 1350g

    = 1.35kg


    Hope this helps
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Dylann)
    You did all the hard ones, the last one is probably the easiest!

    First work out how many moles of Ca(OH)2 you need to make.

    Use Mass = Mr x Mol, where mass = 1kg

    Spoiler:
    Show
    1000 = (40.1 + 34) x mol

    mol = 1000/(40.1+34)

    mol = 13.4953...


    Use this answer to work out the moles of Limestone needed

    Spoiler:
    Show
    Everything relevant uses a 1:1 mole ratio, so there is 13.4953 moles of Limestone required.


    Now using moles of Limestone to work out mass

    Spoiler:
    Show
    mass = mr x mol

    mass = 13.4953... x 100.1
    = 1350g

    = 1.35kg


    Hope this helps

    Thank you so much! haha I was totally overcomplicating it
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: October 12, 2014

University open days

  • Manchester Metropolitan University
    Postgraduate Open Day Postgraduate
    Wed, 14 Nov '18
  • University of Chester
    Chester campuses Undergraduate
    Wed, 14 Nov '18
  • Anglia Ruskin University
    Ambitious, driven, developing your career & employability? Aspiring in your field, up-skilling after a career break? Then our Postgrad Open Evening is for you. Postgraduate
    Wed, 14 Nov '18
Poll
Should Banksy be put in prison?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.