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Topic: Stoichiometry  (Read 14130 times)

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Lialani

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Stoichiometry
« on: September 27, 2005, 07:21:48 PM »
A 20.0 ML sample of 0.200-molar K2CO3 solution is added to 30.0 ML of 0.400-molar Ba(NO3)2 solution. A precipitate occurs. Determine the concentration of barium ion, Ba2+ in solution after the reaction.

Here's my work:

Balance Equation: K+ + CO32- + Ba2+ + NO3- ---> BaCO3

.020L K2CO3(.200M K2CO3) = .004mol K2CO3

.004mol K2CO3(198.2g/mol) = 0.5528g K2CO3

.030L Ba(NO3)2(.400M Ba(NO3)2) = .012 mol Ba(NO3)2

.012mol Ba(NO3)2(261.3g/mol) = 3.1356g Ba(NO3)2

.004mol K2CO3(1 K2CO3 / 1 Ba(NO3)2)(261.3g Ba(NO3)2 / mol) = 1.0452 g Ba(NO3)2 needed

therefore K2CO3 is limitor

and then i'm stuck...
what should i do next?
should i take the mass of the CO3? or should i take the #of mols that i also calculated out for Ba2+ and divide that by .050L ?
Is what i already did right?  ???
help please. thanx.
« Last Edit: September 27, 2005, 07:23:59 PM by Lialani »
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Offline Donaldson Tan

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Re:Stoichiometry
« Reply #1 on: September 27, 2005, 08:45:33 PM »
since K2CO3 is the limiting reagent, how many moles of barium carbonate can be formed from 20.0ML of 0.200M K2CO3?
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Offline constant thinker

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Re:Stoichiometry
« Reply #2 on: September 27, 2005, 08:53:43 PM »
This kinda takes the work out of things but http://theodoregray.com/PeriodicTable/MSP/BalanceReactions
I found that. It's really useful. It elimates the human element to make mistakes. Although it should only be used to check against your answers.
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Lialani

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Re:Stoichiometry
« Reply #3 on: September 27, 2005, 11:13:12 PM »
Quote from: geodome on September 27, 2005, 08:45:33 PM
since K2CO3 is the limiting reagent, how many moles of barium carbonate can be formed from 20.0ML of 0.200M K2CO3?

so...

.004mol K2CO3 (1 mol BaCO3 / 1 mol K2CO3) = .004 mol BaCO3

and since the total L of the solution is .050L is it just .004 mol BaCO3
 divided by .050 L?
« Last Edit: September 27, 2005, 11:18:09 PM by Lialani »
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