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    Question:

    The chloride of an element Z reacts with water according to the following equation.

    ZCL(4) (l) + 2H(2)O(l) > ZO(2) (s) + 4HCL(aq)

    A 1.304g sample of ZCL(4) was added to water.

    The solid ZO(2) was removed by filtration and the resulting solution was made up to 250cm(cubed) in a volumentric flask. A 25.0cm(cubed) portion of this solution
    was titrated against a 0.112 M(molarity) of Sodium Hydroxide, of which 21.7cm(cubed) were required to reach the end point.

    -----------------------------------------------------------------------------------------------------------------------------------------------------------------
    -------------------------------------------------------------------------------------------------------------------------------------------------------------------------
    Use this information to calculate the number of moles of HCL produced and hence the number of moles of ZCL(4) present in the sample.

    Calculate the relative molecular masss, Mr, of ZCL(4).

    From your answer deduce the relative atomic mass of, Ar, of element Z and hence its identity.
    :confused:

    I've tried everything and i keep getting astronomical number for the Mr of ZCL(4)
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    Firstly, excessive bumping (8 times in an hour, give us a fricking BREAK!) on CHRISTMAS DAY is only likely to PISS PEOPLE OFF. It cant be so urgent that you need to pester people on one of the few days of the year people dont want to worry about work!

    Oh, and it's Cl not CL, and H[*sub]2[*/sub]O gives you H2O and cm[*sup]3[*/sup] gives you cm3 (deleting the hashes).

    (Original post by TobeTheHero)
    Question:

    The chloride of an element Z reacts with water according to the following equation.

    ZCL(4) (l) + 2H(2)O(l) > ZO(2) (s) + 4HCL(aq)
    The solid ZO(2) was removed by filtration and the resulting solution was made up to 250cm(cubed) in a volumentric flask. A 25.0cm(cubed) portion of this solution
    was titrated against a 0.112 M(molarity) of Sodium Hydroxide, of which 21.7cm(cubed) were required to reach the end point. [/QUOTE]

    Steps:
    Write an equation for the reaction of HCl and NaOH.
    How much HCl is present in 25cm3 to neutralise 21.7cm3 of 0.112M NaOH?
    How much HCl is present in 250cm3?
    How much ZCl4 will be needed to make this amount of HCl?
    What is the Mr of the compound if that number of moles has a mass of 1.304g?
    What will the Ar of Z be then?

    Show us your calculations and we might be able to diagnose the problem. My bet is you missed out step number 3.
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    Why are you doing homework on Christmas Day?
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    (Original post by gingerbreadman85)
    Firstly, excessive bumping (5 times in an hour, give us a fricking BREAK!) on CHRISTMAS DAY is only likely to PISS PEOPLE OFF. It cant be so urgent that you need to pester people on one of the few days of the year people dont want to worry about work!

    Oh, and it's Cl not CL, and H[*sub]2[*/sub]O gives you H2O and cm[*sup]3[*/sup] gives you cm3 (deleting the hashes).



    The solid ZO(2) was removed by filtration and the resulting solution was made up to 250cm(cubed) in a volumentric flask. A 25.0cm(cubed) portion of this solution
    was titrated against a 0.112 M(molarity) of Sodium Hydroxide, of which 21.7cm(cubed) were required to reach the end point.
    Steps:
    Write an equation for the reaction of HCl and NaOH.
    How much HCl is present to neutralise 21.7cm3 of 0.112M NaOH?
    How much ZCl4 will be needed to make this amount of HCl?
    What is the Mr of the compound if that number of moles has a mass of 1.304g?
    What will the Ar of Z be then?

    Show us your calculations and we might be able to diagnose the problem.[/QUOTE]

    Sorry man
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    (Original post by chembob)
    Why are you doing homework on Christmas Day?
    I don't celebrate xmas I don't believe in christmas

    & I've exams in Jan 2011!
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    (Original post by TobeTheHero)
    I don't celebrate xmas I don't believe in christmas

    & I've exams in Jan 2011!
    Yeah, you and just about every other yr11, 12 and 13 studying Chemistry, that still doesn't entitle you to badger other people on a public holiday. FFS, 8 bumps in an hour, learn some patience!

    Or some respect for the fact you are asking REAL people to spend time helping you and perhaps you owe them some consideration and respect for that rather than nagging them like they owe you something.

    It is possible to ask for people's help without being an insufferable pain in the backside about it. I suggest you learn rather quickly if you expect people to help you out of the goodness of their hearts rather than telling you which cliff to walk off.
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    (Original post by gingerbreadman85)
    Yeah, you and just about every other yr11, 12 and 13 studying Chemistry, that still doesn't entitle you to badger other people on a public holiday. FFS, 8 bumps in an hour, learn some patience!

    Or some respect for the fact you are asking REAL people to spend time helping you and perhaps you owe them some consideration and respect for that rather than nagging them like they owe you something.

    It is possible to ask for people's help without being an insufferable pain in the backside about it. I suggest you learn rather quickly if you expect people to help you out of the goodness of their hearts rather than telling you which cliff to walk off.
    dude chill out!!

    bad christmas or what?

    let the guy bump his thread 8 times and recieve no response. I also saw this thread earlier just as i logged on to TSR and it was too long to answer on christmas so left it. Think you should do the same, until tomorrow
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    (Original post by gingerbreadman85)
    Yeah, you and just about every other yr11, 12 and 13 studying Chemistry, that still doesn't entitle you to badger other people on a public holiday. FFS, 8 bumps in an hour, learn some patience!

    Or some respect for the fact you are asking REAL people to spend time helping you and perhaps you owe them some consideration and respect for that rather than nagging them like they owe you something.

    It is possible to ask for people's help without being an insufferable pain in the backside about it. I suggest you learn rather quickly if you expect people to help you out of the goodness of their hearts rather than telling you which cliff to walk off.
    I sincerely apologise... I should have been more thoughtful
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    (Original post by ?!master?!mini?!)
    dude chill out!!

    bad christmas or what?

    let the guy bump his thread 8 times and recieve no response. I also saw this thread earlier just as i logged on to TSR and it was too long to answer on christmas so left it. Think you should do the same, until tomorrow
    Thank you I hope you had a wonderful xmas day
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    (Original post by TobeTheHero)
    Question:

    The chloride of an element Z reacts with water according to the following equation.

    ZCL(4) (l) + 2H(2)O(l) > ZO(2) (s) + 4HCL(aq)

    A 1.304g sampl of ZCL(4) was added to water.

    The solid ZO(2) was removed by filtration and the resulting solution was made up to 250cm(cubed) in a volumentric flask. A 25.0cm(cubed) portion of this solution
    was titrated against a 0.112 M(molarity) of Sodium Hydroxide, of which 21.7cm(cubed) were required to reach the end point.

    Use this information to calculate the number of moles of HCL produced and hence the number of moles of ZCL(4) present in the sample.

    Calculate the relative molecular masss, Mr, of ZCL(4).

    From your answer deduce the relative atomic mass of, Ar, of element Z and hence its identity.
    :confused:

    I've tried everything and i keep getting astronomical number for the Mr of ZCL(4)
    Couldnt be bothered reading the rest of the thread if you still dont have the answer in the morning (time of this post +9 hours)quote me and I'll try help. I'm tired now sorry.
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    (Original post by gingerbreadman85)
    Firstly, excessive bumping (8 times in an hour, give us a fricking BREAK!) on CHRISTMAS DAY is only likely to PISS PEOPLE OFF. It cant be so urgent that you need to pester people on one of the few days of the year people dont want to worry about work!

    Oh, and it's Cl not CL, and H[*sub]2[*/sub]O gives you H2O and cm[*sup]3[*/sup] gives you cm3 (deleting the hashes).



    The solid ZO(2) was removed by filtration and the resulting solution was made up to 250cm(cubed) in a volumentric flask. A 25.0cm(cubed) portion of this solution
    was titrated against a 0.112 M(molarity) of Sodium Hydroxide, of which 21.7cm(cubed) were required to reach the end point.
    Steps:
    Write an equation for the reaction of HCl and NaOH.
    How much HCl is present in 25cm3 to neutralise 21.7cm3 of 0.112M NaOH?
    How much HCl is present in 250cm3?
    How much ZCl4 will be needed to make this amount of HCl?
    What is the Mr of the compound if that number of moles has a mass of 1.304g?
    What will the Ar of Z be then?

    Show us your calculations and we might be able to diagnose the problem. My bet is you missed out step number 3.[/QUOTE]

    where did you get NaOH from this question is freaking me out!
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    Sodium Hydroxide = NaOH, this is used to neutralise the HCl, simple Acid/Base reaction.

    That your teacher has given you this question assumes that you know how to write an equation for this and do titration calculations to calculate the number of moles of HCl.

    After that point, the rest is just simple moles calculations that you should be able to work through pretty easily.

    Fundamentally this is not to difficult a question. Break it down, take it slow, and know your equations.

    no moles = mass/Mr

    concentration = no moles / volume (dm3)

    1000cm3 = 1dm3

    As i said, if you have issues, show us the calculations you have done which will allow us to work out where you went wrong. The obvious step which would give you an"astronomical" Ar would be to miss that the HCl has been made up to 250cm3 of which only 25cm3 has been titrated with NaOH. Otherwise you are either using the wrong equations or messed up the working.

    The answer gives you an atom which does have the ability to covalently bond to 4 Chlorine atoms. There aren't many of those. Bit of an oddity actually, but that's by the by.
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    (Original post by gingerbreadman85)
    Sodium Hydroxide = NaOH, this is used to neutralise the HCl, simple Acid/Base reaction.

    That your teacher has given you this question assumes that you know how to write an equation for this and do titration calculations to calculate the number of moles of HCl.

    After that point, the rest is just simple moles calculations that you should be able to work through pretty easily.

    Fundamentally this is not to difficult a question. Break it down, take it slow, and know your equations.

    no moles = mass/Mr

    concentration = no moles / volume (dm3)

    1000cm3 = 1dm3

    As i said, if you have issues, show us the calculations you have done which will allow us to work out where you went wrong. The obvious step which would give you an"astronomical" Ar would be to miss that the HCl has been made up to 250cm3 of which only 25cm3 has been titrated with NaOH
    :rofl: I can't believe I missed the first step!? I overlooked sodium hydroxide

    I think I understand now I'll post my working and correct me if I'm wrong
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    (Original post by gingerbreadman85)
    Firstly, excessive bumping (8 times in an hour, give us a fricking BREAK!) on CHRISTMAS DAY is only likely to PISS PEOPLE OFF. It cant be so urgent that you need to pester people on one of the few days of the year people dont want to worry about work!

    Oh, and it's Cl not CL, and H[*sub]2[*/sub]O gives you H2O and cm[*sup]3[*/sup] gives you cm3 (deleting the hashes).



    The solid ZO(2) was removed by filtration and the resulting solution was made up to 250cm(cubed) in a volumentric flask. A 25.0cm(cubed) portion of this solution
    was titrated against a 0.112 M(molarity) of Sodium Hydroxide, of which 21.7cm(cubed) were required to reach the end point.
    (Original post by gingerbreadman85)
    Steps:
    Write an equation for the reaction of HCl and NaOH.
    How much HCl is present in 25cm3 to neutralise 21.7cm3 of 0.112M NaOH?
    How much HCl is present in 250cm3?
    How much ZCl4 will be needed to make this amount of HCl?
    What is the Mr of the compound if that number of moles has a mass of 1.304g?
    What will the Ar of Z be then?

    Show us your calculations and we might be able to diagnose the problem. My bet is you missed out step number 3.
    a) HCl + NaOH (arrow) NaCl + H2O

    b) n(HCl) = 0.112 x 21.7/ 1000 = 0.00243 mol

    c) amount = 0.112 x 250/1000 = 0.028 mol

    d) ???

    stuck

    e) ???
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    (Original post by TobeTheHero)
    c) amount = 0.112 x 250/1000 = 0.028 mol
    What are you doing here? I dont get why you are doing this particular calculation. You have x moles in 25cm3 how many will there be in 250cm3?

    (Original post by TobeTheHero)
    d) ???
    e) ???
    How many moles of HCl are produced from 1 mole of ZCl4..... thus how many moles of ZCl4 will you need to make y amount of HCl?
    • Thread Starter
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    (Original post by gingerbreadman85)
    What are you doing here? I dont get why you are doing this particular calculation. You have x moles in 25cm3 how many will there be in 250cm3?



    How many moles of HCl are produced from 1 mole of ZCl4..... thus how many moles of ZCl4 will you need to make y amount of HCl?
    please help! I don't understand

    can show me your workings? then I can check where I went wrong

    edit:

    if 25 contains n moles, then 250cm^3 contains 250 times n/ 25 ?? am I right?
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    (Original post by TobeTheHero)
    if 25 contains n moles, then 250cm^3 contains 250 times n/ 25 ?? am I right?
    so do that to your answer for b)
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    Please someone post workings! I still don't get it
    this is the only problem i have in AS chemistry MOLES!
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    You've already done all the hard work
    (Original post by TobeTheHero)
    a) HCl + NaOH (arrow) NaCl + H2O

    b) n(HCl) = 0.112 x 21.7/ 1000 = 0.00243 mol

    c) ......

    (Original post by TobeTheHero)
    if 25 contains n moles, then 250cm^3 contains 250 times n/ 25 ?? am I right?
    So you can work out how much HCl is present in 250cm3?
    Then:
    How many moles of ZCl4 will be needed to make this amount of HCl?
    What is the Mr of the ZCl4 if that number of moles has a mass of 1.304g?
    What will the Ar of Z be then?
    • Thread Starter
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    (Original post by gingerbreadman85)
    You've already done all the hard work





    So you can work out how much HCl is present in 250cm3?
    Then:
    How many moles of ZCl4 will be needed to make this amount of HCl?
    What is the Mr of the ZCl4 if that number of moles has a mass of 1.304g?
    What will the Ar of Z be then?
    I'm still confused?! do i use the answer to b) to work out how much hcl is present in 250?
    but i worked how much hcl is present in 21.7cm^3 NOT 25?!
 
 
 
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