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    Hey! Can anyone help me with the following question ?

    in an experiment, an 8.14 g sample of lead reacted completely with a 2.00 mol dm^-3 solution of nitric acid.
    calculate the volume in dm^3, of nitric acid required for the complete reaction.

    equation - 3Pb(s) +8HNO3(aq) -----> 3Pb(NO3)2(aq) + 2NO(g) + 4H2O(l)

    all help and hints are appreciated!!!
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    Is it 0.05?
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    (Original post by AnALevelStudent)
    Is it 0.05?
    i dont know i was getting so confused and i couldnt even get a method that looked right.....
    what was your method???
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    Moles (n) = Mass/Mr, in this case, considering Lead, with a Mr of 207.2, n = 8.14/207.2, which is roughly 0.039 moles

    In this reaction, it is a 3:8 ratio, thus, to find the amount of moles that Nitric Acid reacted with, you divide 0.039 by 3, and multiply by 8, which gives you 0.104.

    You should know that Concentration is equal to n/v (where n is moles and v is volume, in dm^3)
    rearranging this, we should get cv = n, thus v, the variable we're looking for, is equal to n/c
    Substituting our 2 known values of n and c, 0.104 and 2.00 respectively, we get
    0.104/2 = 0.052 (i rounded this up but around this point) dm^3.
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    (Original post by tats bentata)
    i dont know i was getting so confused and i couldnt even get a method that looked right.....
    what was your method???
    Hope this makes sense
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    Yeah I meant 0.052 idk why I put 0.055
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    thank you both so much!!!
    i think i started correctly but then forgot about the ratio....
    very much appreciated!!!
 
 
 
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