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Topic: Limiting Reagent Problem combined with Gas Law Problem  (Read 9160 times)

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Offline nightowl1

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Limiting Reagent Problem combined with Gas Law Problem
« on: November 04, 2009, 09:53:59 PM »
Calcium Carbonate reacts with hydrochloric acid to produce carbon dioxide gas.  If 35.3 g of calcium carbonate reacts with 100ml of 6.00 M HCL, how many liters of carbon dioxide gas will be produced at 745mmHg and 23.0Celcius.

Can someone please walk me through this.
I haven't done a limiting reagent problem for about 2 months.

Ca(CO3) + HCL ----->  CaCl2 + H2(CO3)
I'm not sure if I have to put CO2 at the end of the equation or not.

The answer to the problem is 7.43L

Offline nj_bartel

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Re: Limiting Reagent Problem combined with Gas Law Problem
« Reply #1 on: November 04, 2009, 10:07:37 PM »
Carbonic acid decomposes to water and carbon dioxide.

First, balance your equation.  Second, figure out the moles of all of your reagents.  Multiply the moles by their coefficient in the equation.  See which one is smallest (limiting).  Make your calculations using that value.

Offline nightowl1

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Re: Limiting Reagent Problem combined with Gas Law Problem
« Reply #2 on: November 04, 2009, 10:17:00 PM »
So are you saying the equation should look like this?
Ca(CO3) + HCL -----> H2O + CO2

and balanced it will look like:
Ca(CO3) + 2HCL -----> H2O + CO2

Offline nightowl1

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Re: Limiting Reagent Problem combined with Gas Law Problem
« Reply #3 on: November 05, 2009, 12:01:06 AM »
Okay the limiting reagent is Calcium Carbonate @ 0.353mol (I think)
Now, how am I suppsed to calculate the number of moles of Carbon Dioxide with this?

Offline nj_bartel

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Re: Limiting Reagent Problem combined with Gas Law Problem
« Reply #4 on: November 05, 2009, 12:04:13 AM »
So are you saying the equation should look like this?
Ca(CO3) + HCL -----> H2O + CO2

and balanced it will look like:
Ca(CO3) + 2HCL -----> H2O + CO2


Yes

Offline nj_bartel

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Re: Limiting Reagent Problem combined with Gas Law Problem
« Reply #5 on: November 05, 2009, 12:05:44 AM »
Okay the limiting reagent is Calcium Carbonate @ 0.353mol (I think)
Now, how am I suppsed to calculate the number of moles of Carbon Dioxide with this?


I didn't check your calculations, but if you want to show me how you set up to get that I'll check that.

Anyway, let's assume it's correct.  So you have 0.353 mol calcium carbonate.  How many moles of CO2 are produced per mole of calcium carbonate that reacts?

Offline nightowl1

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Re: Limiting Reagent Problem combined with Gas Law Problem
« Reply #6 on: November 05, 2009, 12:08:05 AM »
1 mole.
so mole to mole ratio is one to one.
How do i find moles of CO2?

Wait, so does that mean that moles of Carbon Dioxide is the same as moles of Calcium Carbonate? 0.353

Offline csrscience.com

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Re: Limiting Reagent Problem combined with Gas Law Problem
« Reply #7 on: November 05, 2009, 12:36:24 AM »
Quote
Wait, so does that mean that moles of Carbon Dioxide is the same as moles of Calcium Carbonate? 0.353

Yes it does.. its all about ratios  :)
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Offline nj_bartel

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Re: Limiting Reagent Problem combined with Gas Law Problem
« Reply #8 on: November 05, 2009, 12:49:08 AM »
Yes :)

Oh, and I think it was accidental, but you left out calcium chloride as a product when you balanced your equations.

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