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Topic: Molarity and Chemical Equations  (Read 7226 times)

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Offline Sis290025

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Molarity and Chemical Equations
« on: October 18, 2005, 05:24:37 PM »
A. Determine the total number of moles of solid sodium hydroxide required to react with 486.00mL of 0.8000M phosphoric acid (H3PO4).

Did I set up the equation correctly?
3NaOH + H3PO4 --> Na3PO4 + 3H20

Then, mol H3PO4(3mol NaOH/1mol H3PO4) = mol NaOH, where mol H3PO4 = (0.8000M)(0.486L)?

B. Determine the concentration of 232.7mL of lead nitrate, Pb(NO3)2 if the sample reacts exactly with 241.0mL of 0.391M potassium iodide (KI).


Equation: Pb(N03)2 + 2KI --> PbI2 + 2KNO3??

so mol PB(NO3)2 = 0.391 M KI *(.241 L KI) (1 mol Pb(NO3)2/ 2 mol KI)

Thanks. Please reply.
« Last Edit: October 18, 2005, 05:26:32 PM by Sis290025 »

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Re:Molarity and Chemical Equations
« Reply #1 on: October 18, 2005, 06:08:53 PM »
Equations looks OK. Show final results and I will tell you if they are OK.
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Offline Sis290025

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Re:Molarity and Chemical Equations
« Reply #2 on: October 18, 2005, 06:20:29 PM »
a. 1.17 mol NaOH

b. 0.0471 M Pb(NO3)2

(from set-up)

Thanks.

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Re:Molarity and Chemical Equations
« Reply #3 on: October 18, 2005, 06:52:31 PM »
a. 1.17 mol NaOH

OK

Quote
b. 0.0471 M Pb(NO3)2

Wrong. You forgot one step of calculations.
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Offline Sis290025

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Re:Molarity and Chemical Equations
« Reply #4 on: October 18, 2005, 07:21:01 PM »
For B, that was the mol of Pb(NO3)2.   :o

For M, 0.0471 mol Pb(NO3)2/ .2327 L Pb(NO3)2 = 0.202 M .

Must remember to read question! *ties bow to thumb*  :-[

Thanks.
« Last Edit: October 18, 2005, 07:22:31 PM by Sis290025 »

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Offline Sis290025

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Re:Molarity and Chemical Equations
« Reply #6 on: October 19, 2005, 06:50:50 AM »
Who's Kim?  ???

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Re:Molarity and Chemical Equations
« Reply #7 on: October 19, 2005, 07:08:13 AM »
Author of the moved post.
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