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Offline blakblob

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Molarity
« on: September 17, 2008, 12:42:02 PM »
Any help would be greatly appreciated. These last 3 problems in my homework have me stumped. I understand molarity and mole ratios of a balanced equation, just can't seem to get the process right here. Thank you!

1.
The typical concentration of HCl in stomach acid (gastric juice) is about 0.082 M. A person experiences `acid stomach' when this concentration rises to about 0.100 Molar. A particular antacid tablet contains 318 mg of the active ingredient NaAl(OH)2CO3 with a molecular weight of 144.0 g/mole. The reaction with stomach acid produces NaCl, AlCl3, CO2, and H2O.

Assume that a person has an `acid stomach' that contains 875 mL of gastric juice at 0.100 M. What will be the concentration of HCl in the gastric juice after the addition of one antacid tablet (give your answer in Molarity)? Assume that the volume of liquid remains constant.

Hint: You need the balanced equation from the previous question. Calculate the moles of acid in this stomach, subtract the moles of antacid using the proper mole ratio, divide by the volume of liquid.

2.
If 70.0 mL of a 5.00 e-02 M solution of AgNO3 is mixed with 90.0 mL of 3.50 e-02 M K2CO3, what mass of solid silver carbonate forms?

3.
Magnesium metal reacts with HCl solution, liberating H2 gas and generating Mg2+ cations in solution. A 2.355-g sample of Mg metal is added to 100.0 mL of a 4.00 M HCl solution, and the reaction goes to completion. What mass of H2 is formed?
« Last Edit: September 17, 2008, 01:07:31 PM by blakblob »
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Offline Borek

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Re: NEED HELP WITH 3 PROBLEMS!!!
« Reply #1 on: September 17, 2008, 12:49:37 PM »
Please read forum rules.

Why do I feel like I am repeating myself.
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Offline blakblob

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Re: NEED HELP WITH 3 PROBLEMS!!!
« Reply #2 on: September 17, 2008, 01:10:15 PM »
Quote from: Borek on September 17, 2008, 12:49:37 PM
Please read forum rules.

Why do I feel like I am repeating myself.

Sorry! Hope I fixed the problem.
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Offline Borek

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Re: Molarity
« Reply #3 on: September 17, 2008, 02:07:01 PM »
Better now, although these are all just stoichiometric questions. But you still have to try. All three questions should be started with writing balanced reaction equation.
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Offline Arkcon

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Re: NEED HELP WITH 3 PROBLEMS!!!
« Reply #4 on: September 17, 2008, 04:58:08 PM »
Quote from: Borek on September 17, 2008, 12:49:37 PM
Please read forum rules.

Why do I feel like I am repeating myself.

September. ;D
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Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline blakblob

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Re: Molarity
« Reply #5 on: September 17, 2008, 06:23:28 PM »
I have got the balanced equations.

For the first problem I posted I get: NaAl(OH)2CO3 + 4HCl --> NaCl + AlCl3 + CO2 +3H2O

From there I get the moles of HCl present in the stomach before the tablet is added: .1M * .875L = .0875mol HCL

Then the moles of the the tablet: 3.19g HCl * 1mol HCl/36.461g HCl * 1mol NaAl(OH)2CO3/4mol HCl * 144g NaAl(OH)2CO3/1mol NaAl(OH)2CO3 = 3.149g NaAl(OH)2CO3/144g/mol NaAl(OH)2CO3 = .0219mol NaAl(OH)2CO3

Then Subtract moles of HCl from moles of NaAl(OH)2CO3 and divide by the number of liters of solution: .0875mol - .0219mol = .0656mol/.875L = .075M

However when I plug this answer into the my online homework it says it is wrong.

As far as the other two problems go I am trying similar processes but cannot come up with the right answer. Maybe I am getting the equations wrong. I don't know.
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Offline macman104

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Re: Molarity
« Reply #6 on: September 17, 2008, 07:13:05 PM »
Quote from: blakblob on September 17, 2008, 06:23:28 PM
I have got the balanced equations.

For the first problem I posted I get: NaAl(OH)2CO3 + 4HCl --> NaCl + AlCl3 + CO2 +3H2O

From there I get the moles of HCl present in the stomach before the tablet is added: .1M * .875L = .0875mol HCL
Good and good.  This is where your problem comes in...
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Then the moles of the the tablet: 3.19g HCl * 1mol HCl/36.461g HCl * 1mol NaAl(OH)2CO3/4mol HCl * 144g NaAl(OH)2CO3/1mol NaAl(OH)2CO3 = 3.149g NaAl(OH)2CO3/144g/mol NaAl(OH)2CO3 = .0219mol NaAl(OH)2CO3
You are not calculating the moles of the antacid in the tablet here.  You are calculating how many moles of NaAl(OH)2CO3 are required to react with all of the HCl.

You need to calculate how many moles of NaAl(OH)2CO3 are present in one antacid tablet (because that's how much they tell you the person takes), and then find out how many moles of HCl react with the antacid.  Then you figure out how many moles you have left and calculate molarity.
Quote
As far as the other two problems go I am trying similar processes but cannot come up with the right answer. Maybe I am getting the equations wrong. I don't know.
Right, so the reason you are asked to post how you are working out the problem is partially so we aren't doing homework for people, but more importantly because it allows us to see where you are making mistakes.  If you post how you were solving the other problems, it will be easier for someone to help you figure out where you're going wrong (if you are wrong).
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Offline blakblob

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Re: Molarity
« Reply #7 on: September 17, 2008, 08:35:35 PM »
Thank You! I managed to get the first one.

For the second one I am starting with the equation: AgNO3 + K2CO3 --> AgCO3 + K2NO3 (not sure what silver carbonate compound is) Kind of lost after that. Hoping for a little direction.

For the third I am starting with the equation: 2Mg + 2HCl --> H2 + 2MgCl

I find the mass of HCl present: 4M * .1L = .4mol * 36.461g/mol = 14.584g

Then because Mg is the limiting reactant: 2.355g Mg * 1mol Mg/24.305g Mg * 1mol H2/2mol Mg * 2.016g H2/1mol H2 = .098g H2
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Offline macman104

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Re: Molarity
« Reply #8 on: September 17, 2008, 09:26:15 PM »
Quote from: blakblob on September 17, 2008, 08:35:35 PM
For the second one I am starting with the equation: AgNO3 + K2CO3 --> AgCO3 + K2NO3 (not sure what silver carbonate compound is) Kind of lost after that. Hoping for a little direction.
Not quite, Silver Carbonate is Ag2CO3, and it should be KNO3.  Balance that, and this is a limiting reagent problem.
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Offline blakblob

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Re: Molarity
« Reply #9 on: September 17, 2008, 11:04:20 PM »
Ok I did:

2.06e-5 g AgNO3 * 1 mol AgNO3/169.873 g AgNO3 * 1 mol Ag2CO3/2 mol AgNO3 * 275.746 g Ag2CO3/1 mol Ag2CO3 = 1.67e-5 g Ag2CO3

It said that was wrong. Do you see my mistake? Trying to get this solved before 11 central time. Thanks again.
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Offline macman104

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Re: Molarity
« Reply #10 on: September 17, 2008, 11:42:02 PM »
You need to determine whether AgNO3 or K2CO3 are the limiting reagent.  You need to see which one makes less Ag2CO3, and that will be your answer.

Read through this:

http://dbhs.wvusd.k12.ca.us/webdocs/Stoichiometry/Limiting-Reagent.html
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Offline blakblob

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Re: Molarity
« Reply #11 on: September 17, 2008, 11:51:08 PM »
I have tried both. Neither of them work.

If you could please show me the process you would use I would greatly appreciate it. I have been working on these problems for almost 6 hours now. I am not wanting free answers, but I have tried everything that my tiny brain can think of to make this problem work. lol
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Offline Borek

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Re: Molarity
« Reply #12 on: September 18, 2008, 03:10:47 AM »
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Offline blakblob

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Re: Molarity
« Reply #13 on: September 18, 2008, 11:10:00 AM »
Thank you all. I finally figured out what I was doing wrong in the end.
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