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-   -   Easy Questions 1)find the maximum possible amount 2)Electrical resistance (https://www.askmehelpdesk.com/showthread.php?t=546452)

  • Jan 21, 2011, 10:42 AM
    pop000
    Easy Questions 1)find the maximum possible amount 2)Electrical resistance
    1) I need to find the maximum possible amount of PbI2, that Created in reaction between 0.3 mol Pb(NO3)2 to 0.2mol KI

    Here is 4 answers Options:

    A)0.1 mol

    B)0.15mol

    C)0.3 mol

    D)0.5 mol

    2) tell me if I correct , in this following list the most good Electrical resistance Compound in State of matter Liquid us AlF3 Because there is a Ionic bond YES?
    A)CF4
    B)AlF3
    C)SF2
    D)HF
    Thanks :)
  • Jan 21, 2011, 10:53 AM
    Unknown008

    1)

    Write the balanced chemical equation.
    Find the mole ratio of Pb(NO3)2 to KI
    From this, determine the limiting reagent.
    Find the mole ratio of the limiting reagent to PbI2.
    Use the number of moles of the limiting reagent and the mole ratio you just found to get the maximum amount of moles of PbI2 that you can get.

    2) If AlF3 is in the molten state, then yes, it is the most conducting substance :)
  • Jan 22, 2011, 04:17 AM
    pop000
    hi. Thanks for answer but, I need help with to Write the balanced chemical equation.

    I know that the basis equation is something like this: Pb(NO3)2+KI---->PbI2 but how I do the balanced ?
    how do I know what Coefficient need to be in Which Material, if I was know any Coefficient is was make it Easier :)


    for 2) yes is in molten state.

    thanks.
  • Jan 22, 2011, 05:25 AM
    Unknown008

    Okay, count each type of atom on each side of the equation.

    There are
    Pb - ? Ions
    NO3 - ? Molecular ions
    K - ? Ions
    I - ? Ions

    Count those on the left, and those on the right, and you get:
    Left:
    Pb - 1 ion
    NO3 - 2 molecular ions
    K - 1 ions
    I - 1 ions

    Right:
    Pb - 1 ions
    NO3 - 0 molecular ions
    K - 0 ions
    I - 2 ions

    You see what problem there is? You forgot K and NO3. Since Pb 'stuck' to I, K will go with NO3.

    Then, you need 2 I on the left, hence put a 2 as coefficient of KI and let's see what happens.



    Count again. This time, you have one NO3 ion less on the right, and 1 K less on the right. So, put a 2 as coefficient in front of KNO3:



    All is right now. Can you continue now? :)
  • Jan 22, 2011, 07:06 AM
    pop000
    Thanks for answer.

    Well the mole ratio of Pb(NO3)2 to KI 1:2

    But I am not got it how to continue . There is no any Formula to use? I not need to find m and Mm?

    I know is hard to Explain to me ;)

    Thank you
  • Jan 22, 2011, 07:54 AM
    Unknown008

    No, don't worry, if you don't know, we'll try to help you :)

    Ok, from the mole ratio, you know that:
    1 mol Pb(NO3)2 will react with exactly 2 moles of KI

    Let's take Pb(NO3)2 itself, we have 0.3 mole of it.

    How many moles of KI would this react with?
    But then, how many moles of KI is there?

    Hence, which one is there in excess and which one is limiting?

    This is enough for now, but if you can , read my first and answer and try the last two steps :)
  • Jan 22, 2011, 01:52 PM
    pop000
    KI is the one is limiting.

    so all the KI was react mean all the o.2 mol

    and because the ratio of the mole between KI to Pb(NO3)2 is 2:1
    so to know how many mole of Pb(NO3)2 is react with the KI I did 0.2/2=0.1 and this is the answer.

    do I correct ?

    thank you.
  • Jan 23, 2011, 05:10 AM
    Unknown008

    Yes, it is, now, how many moles of PbI2 do you obtain from 0.2 mol of KI, or from 0.1 mol of Pb(NO3)2?

    (of course, you need to set up another mole ratio)
  • Jan 23, 2011, 06:42 AM
    pop000
    Comment on Unknown008's post
    Also 0.1 Because the ratio between PbI2 and Pb(NO3)2 1:1 yes?
  • Jan 23, 2011, 06:52 AM
    Unknown008

    Yes! You got it right. Well done :)
  • Jan 23, 2011, 08:48 AM
    pop000
    Comment on Unknown008's post
    :) thank you without your help I not sure if I was got it.

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