Sciencemadness Discussion Board
Not logged in [Login ]
Go To Bottom

Printable Version  
Author: Subject: Chemistry problem (a championship one)
Avathacis
Harmless
*




Posts: 2
Registered: 20-4-2010
Member Is Offline

Mood: No Mood

[*] posted on 20-4-2010 at 09:00
Chemistry problem (a championship one)


Inorganic chemistry warning, evacuate the building!
I know this is my first post and i am asking for help, but please bear with me.
So yeah here's the problem:
In one litre of a solution there is 0.1 mole of Pb(NO3)2. Into 300 ml of the solution there is a 1.3g mass chunk of Zinc put in. How many moles (or mol's or w/e) of Pb(NO3)2 is left in the solution? (Not quite sure if it asks for the 300 ml part or the one left with 700 ml, might as well just count both.) How many grams of Pb is there after the reaction?
View user's profile View All Posts By User
bbartlog
International Hazard
*****




Posts: 1139
Registered: 27-8-2009
Location: Unmoored in time
Member Is Offline

Mood: No Mood

[*] posted on 20-4-2010 at 09:49


So in 300 ml of this liter solution there will be 300/1000 * 0.1 = 0.03 moles of Pb(NO3)2.
To this is now added 1.3/65.4 = 0.02 moles of Zn.
This will undergo a reaction, Pb(NO3)2 + Zn -> Zn(NO3)2 + Pb, as suggested by the answers the question requires.
Since zinc and lead have the same valence here, 0.02 moles of zinc simply trade places with 0.02 moles of lead,
which leave 0.01 moles of Pb(NO3)2 in solution, and 0.02 moles of lead as precipitate. The atomic weight of lead is 207, so this will be 4.14 grams of lead.
The remaining 700ml of solution (which I think is not likely to be the subject of the question) of course has 0.07 moles of Pb(NO3)2 left in it.

I don't know what 'championship' this is for but this is not a very difficult question.
View user's profile View All Posts By User
Avathacis
Harmless
*




Posts: 2
Registered: 20-4-2010
Member Is Offline

Mood: No Mood

[*] posted on 20-4-2010 at 09:57


Quote: Originally posted by bbartlog  
So in 300 ml of this liter solution there will be 300/1000 * 0.1 = 0.03 moles of Pb(NO3)2.
To this is now added 1.3/65.4 = 0.02 moles of Zn.
This will undergo a reaction, Pb(NO3)2 + Zn -> Zn(NO3)2 + Pb, as suggested by the answers the question requires.
Since zinc and lead have the same valence here, 0.02 moles of zinc simply trade places with 0.02 moles of lead,
which leave 0.01 moles of Pb(NO3)2 in solution, and 0.02 moles of lead as precipitate. The atomic weight of lead is 207, so this will be 4.14 grams of lead.
The remaining 700ml of solution (which I think is not likely to be the subject of the question) of course has 0.07 moles of Pb(NO3)2 left in it.

I don't know what 'championship' this is for but this is not a very difficult question.

9th grade. Dunno, my teacher ripped it out of somewhere. There's not much to give in 9th grade (in our country chemistry starts at 8th and it's just general, a quick slide over the top.)

[Edited on 20-4-2010 by Avathacis]

[Edited on 20-4-2010 by Avathacis]
View user's profile View All Posts By User

  Go To Top