Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    14
    ReputationRep:
    Can anyone help me with gas calculations? I'm struggling with trying to convert moles to mass in relation to gases, do you just do the normal m=NXM. I was trying to do this for question d in the analysis section and I got lost..

    http://www.ocr.org.uk/Images/67673-u...k-specimen.pdf
    • Community Assistant
    • Study Helper
    Offline

    14
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by kath_igidbashian)
    Can anyone help me with gas calculations? I'm struggling with trying to convert moles to mass in relation to gases, do you just do the normal m=NXM. I was trying to do this for question d in the analysis section and I got lost..

    http://www.ocr.org.uk/Images/67673-u...k-specimen.pdf
    Pure substances:

    moles = mass/Mr

    Solutions:

    moles = concentration x volume

    Gases:

    moles = mass/22.4 (at STP)

    BUT, don't forget that gases can also be pure substances, in which case equation 1 also applies.
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by charco)
    Pure substances:

    moles = mass/Mr

    Solutions:

    moles = concentration x volume

    Gases:

    moles = mass/22.4 (at STP)

    BUT, don't forget that gases can also be pure substances, in which case equation 1 also applies.
    Thanks! Is there anyway you could take a look at the question I said and do a work through guide for me? Just to show me the steps in case I get another question like this again?
    • Community Assistant
    • Study Helper
    Offline

    14
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by kath_igidbashian)
    Thanks! Is there anyway you could take a look at the question I said and do a work through guide for me? Just to show me the steps in case I get another question like this again?
    part (d) comes from the answers to (a), (b) and (c).

    What have you got for those?
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by charco)
    part (d) comes from the answers to (a), (b) and (c).

    What have you got for those?
    MgCO3 is 0.00356 mol
    H2SO4 is 0.0005 mol

    H2SO4 is in excess
    • Community Assistant
    • Study Helper
    Offline

    14
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by kath_igidbashian)
    MgCO3 is 0.00356 mol
    H2SO4 is 0.0005 mol

    H2SO4 is in excess
    what about the equation for the reaction?
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by charco)
    what about the equation for the reaction?
    oh sorry
    H2SO4(aq) + MgCO3 (s)-> MgSO4(aq) + H20(l) + CO2(g)
    • Community Assistant
    • Study Helper
    Offline

    14
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by kath_igidbashian)
    oh sorry
    H2SO4(aq) + MgCO3 (s)-> MgSO4(aq) + H20(l) + CO2(g)
    OK, so the equattion tells you that the moles of carbon dioxide will be equal to the moles of the limiting reagent.

    According to your calculations
    MgCO3 is 0.00356 mol
    H2SO4 is 0.0005 mol

    which wuold make magnesium carbonate in excess ...

    But 0.1 x 0.05 moles of sulphuric acid = 0.005 mol NOT 0.0005 mol
    0.3 g of magnesium carbonate = 0.3/84 = 0.00357 mol

    So, yes sulphuric acid is in excess and magnesium carbonate is the limiting reagent.

    Hence moles of CO2 = moles of magnesium carbonate = 0.00357 mol.

    1 mol of gas at RTP = 24000 cm3

    hence 0.00357 mol = 0.00357 * 24000 = 85.7 cm3
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by charco)
    OK, so the equattion tells you that the moles of carbon dioxide will be equal to the moles of the limiting reagent.

    According to your calculations
    MgCO3 is 0.00356 mol
    H2SO4 is 0.0005 mol

    which wuold make magnesium carbonate in excess ...

    But 0.1 x 0.05 moles of sulphuric acid = 0.005 mol NOT 0.0005 mol
    0.3 g of magnesium carbonate = 0.3/84 = 0.00357 mol

    So, yes sulphuric acid is in excess and magnesium carbonate is the limiting reagent.

    Hence moles of CO2 = moles of magnesium carbonate = 0.00357 mol.

    1 mol of gas at RTP = 24000 cm3

    hence 0.00357 mol = 0.00357 * 24000 = 85.7 cm3
    Ah ok, thanks! Is it because of the 1:1 ratio that co2 = MgCO3

    Posted from TSR Mobile
    • Community Assistant
    • Study Helper
    Offline

    14
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by kath_igidbashian)
    Ah ok, thanks! Is it because of the 1:1 ratio that co2 = MgCO3

    Posted from TSR Mobile
    Yes, the carbon dioxide should be equal to the moles of the limiting reagent ...
    • Thread Starter
    Offline

    14
    ReputationRep:
    Thank you so much for your help! You're a life saver!!
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Brexit voters: Do you stand by your vote?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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.