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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: gnats50 on October 06, 2005, 03:50:14 PM

Title: mole calculations
Post by: gnats50 on October 06, 2005, 03:50:14 PM

Hi, this is really simple, but I can't figure it out. It's driving me crazy.

When 50.00 mL of 1.000x10^-1M Pb(NO3)2 solution was added to 50.00 mL of 1.000x10^-1M NaI solution, 1.021 g of PbI2 was obtained as product.
Calculate the number of moles of Pb(NO3)2 used.


Thank You for your help. I really want to know how to get to the answer and not just the answer.

Title: Re:mole calculations
Post by: Borek on October 06, 2005, 04:28:37 PM

Write down reaction equation.

Title: Re:mole calculations
Post by: Mitch on October 06, 2005, 04:29:39 PM

First always start by writing the balanced chemical equation.

Title: Re:mole calculations
Post by: Borek on October 06, 2005, 04:37:01 PM

Quote from: Mitch on October 06, 2005, 04:29:39 PM

First always start by writing the balanced chemical equation.

I was first :)

Title: Re:mole calculations
Post by: gnats50 on October 06, 2005, 04:47:19 PM

Pb(NO3)2(aq)   +   2 NaI(aq)   ----->  2 NaNO3(aq)  +   PbI2(ppt)

Title: Re:mole calculations
Post by: Mitch on October 06, 2005, 05:00:21 PM

You have grams PbI2 convert it to moles PbI2 relate it to moles of Pb(NO3)2 and convert that into grams of Pb(NO3)2 and you're all done. :)