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Topic: stoichiometry calculation  (Read 6018 times)

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lewisheather

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stoichiometry calculation
« on: August 31, 2005, 04:40:32 PM »
I am trying to solve this problem. can you tell me if i did it correctly.

Problem: You have 38.8 grams of K2SO4 and this is dissolved in .1 L of water. What is the molarity of K. MW of K2SO4 is 194.

This is how I solved it:

38.8 g K2SO4/194 g m-1 = .2 moles of K2SO4

.2 moles K2SO4 * 1 mol K / 1 mol K2SO4 = .2 mols of K

the molarity is moles/liter so I divided by .1 L the given amount of fluid.

Hence, .2 molesK/ .1 L = 2M K

Is this correct?
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Offline Borek

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Re:stoichiometry calculation
« Reply #1 on: August 31, 2005, 05:20:09 PM »
Molar mass of K2SO4 is 174, so the final result is wrong - although your approach is correct.

CASC rules :)
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lewisheather

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Re:stoichiometry calculation
« Reply #2 on: August 31, 2005, 10:19:17 PM »
OK that makes sense. These were the exact numbers given to me on the test. I was just worried that I was doing the wrong calculation.

I showed my work so hopefully they will see that they messed up on the numbers. Or maybe they did this to see if we were paying attention.

I might be wrong on the formula of K2SO4, I am going off of memory of what I thought the question of the test was.  

I was just worried that I was doing the set up wrong.

Thanks so much for your help.

Heather
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oldddog

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Re:stoichiometry calculation
« Reply #3 on: September 02, 2005, 02:03:30 AM »
Potassium sulphate dissolves in water according to the equation below;
 K2SO4(s) + H2O ---> 2K+(aq) + SO42-(aq)
So for every mole of potassium sulphate that dissolves you get 2 moles of potassium ions in solution.
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Offline Borek

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Re:stoichiometry calculation
« Reply #4 on: September 02, 2005, 03:39:45 AM »
Oops, did I misread the question again?
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Che-Comp

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Re:stoichiometry calculation
« Reply #5 on: September 07, 2005, 04:36:19 AM »
38.8 g K2SO4 * (1 mol K2SO4/174 g K2SO4) * (2 mols K/1 mol K2SO4)
= 0.446 mols K

Assume volume occupied by K2SO4(aq) is negligible compared to the volume of water.
0.446 mols K/0.1L =  4.46 mols K/1L solution approx. 4M

:)


Quote from: lewisheather on August 31, 2005, 04:40:32 PM
I am trying to solve this problem. can you tell me if i did it correctly.

Problem: You have 38.8 grams of K2SO4 and this is dissolved in .1 L of water. What is the molarity of K. MW of K2SO4 is 194.

This is how I solved it:

38.8 g K2SO4/194 g m-1 = .2 moles of K2SO4

.2 moles K2SO4 * 1 mol K / 1 mol K2SO4 = .2 mols of K

the molarity is moles/liter so I divided by .1 L the given amount of fluid.

Hence, .2 molesK/ .1 L = 2M K

Is this correct?
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