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Stoichiometry help

  1. Mar 7, 2005 #1

    DB

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    A little stoichiometry...
    Here's all that was given:

    A container of HCl spilled on a marble table surface. The marble contains calcium carbonate, which reacts with the acid to produce calcium chloride, carbon dioxide and water.

    a) If 500ml of 0.2 M acid spilled, what mass of calcium carbonate reacted?

    So I wrote out the (combustion is it?) equation.
    [tex]CaCO_3 + 2HCl\rightarrow CaCl_2 + CO_2 + H_2O[/tex]

    Then molar mass:
    [tex]100g(CaCO_3) + 72g(2HCl)\rightarrow 110g(CaCl_2) + 44g(CO_2) +18g( H_2O)[/tex]

    So now I know that 72g of HCl is 2 mol (since 2HCl) so 0.2 mol = 7.2g
    But know I'm stuck...a push in the right direction???

    Also.
    b)How many moles of [itex]CO_2[/itex] were formed during this reaction?
    1 mole
    c)How many molecules of water were formed during this reaction?
    ~6.022*10^23
    I don't think there's any problems there.

    Thanks in advance
     
  2. jcsd
  3. Mar 7, 2005 #2

    dextercioby

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    Yes,there are problems everywhere.So the first.U've computed 0.2 mols of HCl.How many mols of CaCO_{3} will react with it...?

    Daniel.
     
  4. Mar 7, 2005 #3

    saltydog

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    According to the balanced equation, for every mole of HCl reacting, one-half of a mole of [itex]CaCO_3[/itex] reacts. So how many moles of HCl are in 500ml of a 0.2 molar solution? Suppose that's x number of moles. Now, from what I said above, one-half of that number of moles is the number of moles of [itex]CaCO_3[/itex] reacting. Same dif for [itex]CO_2[/itex] right (half as much) and water too for that matter.


    Now, isn't there Avagadro's number of molecules in one mole? So using the infor above (the number of moles of water produced), figure out how many molecules of water.
     
  5. Mar 7, 2005 #4

    Gokul43201

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    DB,

    0.2 M does not mean 0.2 moles. It refers to a solution of strength 0.2 moles per liter. So, how many moles will there be in 500 mL of this solution ?
     
  6. Mar 7, 2005 #5

    dextercioby

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    Ooooooops,didn't see the big M,sorry.

    Daniel.
     
  7. Mar 7, 2005 #6
    Lets start here. You have the concentration and the volume. You can use these to see what the amount of moles is.

    The equation is: [tex] Concentration (mol dm^{-3}) = \frac{Moles (moles)}{Volume (dm^3)}[/tex]

    Use that to find the number of moles of acid that have been spilled. You can then use this to find the mass: [tex]Moles (moles) = \frac{Mass (grams)}{Molar Mass (g mol^{-1})}[/tex]

    You then have the mass of the HCl. Use this you can find the mass of CaCl2 in the normal way, find ratios and molar masses and see what is produced.

    Get this and then I will help with the rest. :smile:

    The Bob (2004 ©)
     
  8. Mar 7, 2005 #7

    DB

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    Ok so, 0.1 moles reacts with [itex]CaCO_3[/itex] in a ratio of 2:1? So, half of 0.1 mol is 0.05 mole which is 5g of [itex]CaCO_3[/itex]

    for b and c, i think you told me that the answers are 0.5 mol and [itex]N_A/2[/itex], but i dont really understand why I'm dividing by 2...
     
  9. Mar 7, 2005 #8
    I got 55.5g of Calcium Chloride. More accurate molar mass of CaCl2 is 111 g mol-1.

    The Bob (2004 ©)
     
  10. Mar 7, 2005 #9

    saltydog

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    Ok.

    No. I think you mean 0.05 moles right? If 0.1 moles of HCl are reacting and the ratio is 2/1 then that means 0.05 moles of water and carbon dioxide are produced right? And if one mole has Avagadros number then one half of a mole would have one half of Avagadros number, one tenth of a mole would have one-tenth of Avagadros number and so if you have 0.05 moles of water then that means you have 0.05 of Avagadros's number of molecules right?
     
  11. Mar 7, 2005 #10

    DB

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    "I got 55.5g of Calcium Chloride. More accurate molar mass of CaCl2 is 111 g mol-1.

    The Bob (2004 ©)"
    calium chloride? for a? the question is asking calcium carbonate
     
  12. Mar 7, 2005 #11
    I apologise. In that case I get 50g of Calcium Carbonate. :smile:

    The Bob (2004 ©)
     
  13. Mar 7, 2005 #12

    DB

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    Naww lol i meant 0.5. im confused. 0.1 moles of acid are reacting we know that.
    So then 0.1 mole of both CO2 and water are...? Im sorry if im being anoying im just really confused.
    :cry:
     
  14. Mar 7, 2005 #13

    DB

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    no prob, just i got 5 g and saltydog said it was ok...
     
  15. Mar 7, 2005 #14
    Here is what I did.

    The amount of moles of the spilled Hydrochloric Acid is 0.2 mol dm-3 x 0.5 dm3 = 0.4 moles.

    Molar Mas of HCl = 36.5 g mol-1

    0.4 moles x 36.5 g mol-1 = 14.6g (now proven)

    The equation you need to look at is the [tex]2HCl + CaCO_3[/tex]

    HCl = 35.6 g mol-1 x 2 moles = 73g
    CaCO3 = 100 g mol-1 x 1 mol = 100g

    Therefore HCl:CaCO3 = 73g:100g

    So 1g of HCl : 1.37g of CaCO3

    Therefore 14.6g of HCl : 1.37 x 14.6 = 20g of CaCO3

    Ok. I have changed my mind because my original answer was wrong because I get the molar mass of HCl wrong (:rofl:) but I am now sure it is 20g.

    The Bob (2004 ©)
     
  16. Mar 7, 2005 #15
    The moles of Carbon Dioxide = 0.2 moles.
    The moles of Water = 0.2 moles as well. I don't know about how many molecules but I personally think it might be the same.

    The Bob (2004 ©)
     
    Last edited: Mar 7, 2005
  17. Mar 7, 2005 #16

    DB

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    Thanks bob, i see the work u've done, i just cant understand how u got 0.4 moles of acid spilled, i always get 0.1....can u explain plz
     
  18. Mar 7, 2005 #17
    [tex]Concentration (mol dm^{-3}) = \frac{Moles (mol)}{Volume (dm^3)}[/tex]

    M (or the molarity) is equal to the same number in mol dm-3 e.g. 0.2M = 0.2 mol dm-3.

    Also the equation needs the amount in dm3. You have it in ml. This can be changed straight away into cm3 e.g. 300 ml = 300 cm3. Then you need it in dm3. As it is a cubic factor then you need to change the number by a cubic number. Let me explain this better.

    1 dm = 10 cm. This is worked out by multiplying the dm by 10 or by dividing the cm by 10.

    1 dm2 = 100 cm2. This time it is worked out by multiplying the dm by 102 (100) or by dividing the cm by 102 (100).

    Therefore 1 dm3 = 1000 cm3 multiplying/dividing by 103 (1000). e.g. 300 ml = 300 cm3 = 0.3 dm3

    The amount of moles of the spilled Hydrochloric Acid is 0.2M x 500 ml = 0.2 mol dm-3 x 500 cm3 = 0.2 mol dm-3 x 0.5 dm3 = 0.1 moles. (Oh man, I have done it again. I need to learn to check what I am doing. Sorry again :frown:).

    The Bob (2004 ©)
     
  19. Mar 7, 2005 #18
    I get 5g as well :frown:. Sorry everyone. I has been good practise for me though and I can see my mistakes clearer and also you can see my method clearly and rather than jumping to conclusion it can be proven (although jumping seemed to work better).

    The Bob (2004 ©)
     
  20. Mar 7, 2005 #19
    It is now:
    The moles of Carbon Dioxide = 0.05 moles.
    The moles of Water = 0.05 moles.

    The Bob (2004 ©)

    P.S. Sorry again :frown: but I need to sleep now. I am really quite tired. Hope what I did helped in some way or another. :smile:
     
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