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    Hi, can someone please help me really stuck on these 2 questions....think I might have figured out the first one but seems kinda wrong.

    A chemist neutralised 25.0 cm^3 0.200 mol. dm^-3 HCL with slaked lime

    Ca(OH)2 + 2HCL -> CaCl2 + 2H2O

    1)how many moles of HCL were neutralised?

    2)Calculate the mass of Ca(OH)2 that neutralises this HCL


    Im thinking for 1:
    mol= (cm/1000) x conc

    So I would have mol= (25/1000) x 0.200 = 0.005

    But then again the question is in mol.dm so I think I might be wrong...

    Can someone please help?

    Cat
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    (Original post by chocoholic_cat)
    Hi, can someone please help me really stuck on these 2 questions....think I might have figured out the first one but seems kinda wrong.

    A chemist neutralised 25.0 cm^3 0.200 mol. dm^-3 HCL with slaked lime

    Ca(OH)2 + 2HCL -> CaCl2 + 2H2O

    1)how many moles of HCL were neutralised?

    2)Calculate the mass of Ca(OH)2 that neutralises this HCL


    Im thinking for 1:
    mol= (cm/1000) x conc

    So I would have mol= (25/1000) x 0.200 = 0.005

    But then again the question is in mol.dm so I think I might be wrong...

    Can someone please help?

    Cat
    This question is seriously easy. You should be ashamed of yourself. It is easy to work out moles of acid, you know the ratio and therefore moles of alkali which were neutalised, then simply apply g=Mr*moles.
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    (Original post by chocoholic_cat)
    Hi, can someone please help me really stuck on these 2 questions....think I might have figured out the first one but seems kinda wrong.

    A chemist neutralised 25.0 cm^3 0.200 mol. dm^-3 HCL with slaked lime

    Ca(OH)2 + 2HCL -> CaCl2 + 2H2O

    1)how many moles of HCL were neutralised?

    2)Calculate the mass of Ca(OH)2 that neutralises this HCL


    Im thinking for 1:
    mol= (cm/1000) x conc

    So I would have mol= (25/1000) x 0.200 = 0.005

    But then again the question is in mol.dm so I think I might be wrong...

    Can someone please help?

    Cat
    i think your answer for the first part is rite.....

    2) 0.005 moles of Calcium hydroxide has reacted....and the RMM of that is 40+(16*2)+2= 74

    moles x RMM = mass
    0.005 x 74 = 0.37 grams
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    (Original post by matouwah)
    i think your answer for the first part is rite.....

    2) 0.005 moles of Calcium hydroxide has reacted....and the RMM of that is 40+(16*2)+2= 74

    moles x RMM = mass
    0.005 x 74 = 0.37 grams
    Thank you *and thank you for not being insulting with your reply too*
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    and the mass of CA(OH)2 is just half that. unless there's some trickery about it. Which there isn't so far as i can see... and i look hard.. and usualy find it where there isn't any...
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    (Original post by Spc_K)
    and the mass of CA(OH)2 is just half that. unless there's some trickery about it. Which there isn't so far as i can see... and i look hard.. and usualy find it where there isn't any...
    yeah....i forgot the ratio was 1:2 for CA(OH)2 : HCL....so the mass is half of what i wrote before..
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    (Original post by matouwah)
    yeah....i forgot the ratio was 1:2 for CA(OH)2 : HCL....so the mass is half of what i wrote before..
    Ok thanks again
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    (Original post by chocoholic_cat)
    Ok thanks again
    Do you still need help with the first question or can you manage it now?
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    (Original post by Chicken)
    Do you still need help with the first question or can you manage it now?
    No thanks Im good, just about to hand it in 2nd lesson, but thanks for the offer
 
 
 
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