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Topic: Na2SO4+Pb(NO3)2 = Pb(SO4)+NaNO3... 0.1M?  (Read 3799 times)

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Offline Nautis

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Na2SO4+Pb(NO3)2 = Pb(SO4)+NaNO3... 0.1M?
« on: May 18, 2013, 12:44:42 AM »
I'm given the following:

Na2SO4+Pb(NO3)2 = Pb(SO4)+NaNO3

Blanced = 1Na2SO4+1Pb(NO3)2 = 1Pb(SO4)+2NaNO3

From this, I must create a .1M of each solution.  I am given a liquid Na2SO4, and solid Pb(NO3)2.
How would I go about doing this?

No more than 25ml of each solution can be created either.
 
What is the theoretical mass of the products?

Offline UG

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Re: Na2SO4+Pb(NO3)2 = Pb(SO4)+NaNO3... 0.1M?
« Reply #1 on: May 18, 2013, 03:01:01 AM »
Were you really given liquid Na2SO4? :o I think it was a more concentrated solution, do you know its concentration? Anyway, say you had 25 mL of a 0.1 M Pb(NO3)2 solution, how many moles of Pb(NO3)2 is in solution? Once you have calculated this, you can find the mass of Pb(NO3)2 needed

Offline Nautis

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Re: Na2SO4+Pb(NO3)2 = Pb(SO4)+NaNO3... 0.1M?
« Reply #2 on: May 18, 2013, 03:14:21 AM »
Thanks for replying :)

It was labeled "Sodium Sulfate 1M"
And i think i may have worded my sentences wrong in the last post.

We were given solid Lead II Nitrate, and a Liquid sodium sulfate labeled "Sodium Sulfate 1M".

We weren't allowed to make more than a 25ml solution for each.

We were to find out how much of each compound would be needed to make a 25ml solution of each.

 

Offline UG

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Re: Na2SO4+Pb(NO3)2 = Pb(SO4)+NaNO3... 0.1M?
« Reply #3 on: May 18, 2013, 03:29:56 AM »
Ok, that's good :) I'm glad that issue is sorted. So then, in reference to my previous post, do you know how to calculate the number of moles of Pb(NO3)2 in 25 mL of a 0.10 M solution? What formulas have you been taught?

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