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Author Topic: How many grams and moles of CO2 gas??  (Read 8484 times)

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Lindsay

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How many grams and moles of CO2 gas??
« on: November 11, 2008, 04:22:22 PM »

Problem 1

A 1.719 g sample of a CaCo3 mixture is thermally decomposed, evolving CO2. After the reaction, the mixture has a mass of 1.048 g

a. How many grams and moles of CO2 gas are evolved?
Answer--
1.719 g - 1.048 = 0.671 g CO2

b. How many moles and grams of CaCO3 are in the mixture?

I'm not sure about this one..

0.0152 mole of CO2 * 1 mol CaCO3/ 1 mol CO2 = 0.0152 mol CaCO3
* 100.09 g CaCO3/ 1 mol CaCO3 = 1.52 g CaCO3

Is the formula for this problem CaCO3 ---- CaO +CO2??

Please advise,

Lindsay
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AWK

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Re: How many grams and moles of CO2 gas??
« Reply #1 on: November 11, 2008, 07:12:06 PM »

Quote
a. How many grams and moles of CO2 gas are evolved?
Answer--
1.719 g - 1.048 = 0.671 g CO2

And moles ?
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Borek

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Re: How many grams and moles of CO2 gas??
« Reply #2 on: November 11, 2008, 08:43:27 PM »

A 1.719 g sample of a CaCo3 mixture

Mixture? What are other components of the mixture? Is there any other source of CO2 between them?
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Lindsay

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Re: How many grams and moles of CO2 gas??
« Reply #3 on: November 12, 2008, 04:14:25 AM »

a. How many grams and moles of CO2 gas are evolved?
Answer--
1.719 g - 1.048 = 0.671 g CO2

0.671 g CO2 * 1 mol CO2/44.01 g = 0.0152 mol of CO2

Mixture?? I have given you the complete problem. I'm thinking maybe CaO is the other one.

Thanks,

Lindsay
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Borek

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Re: How many grams and moles of CO2 gas??
« Reply #4 on: November 12, 2008, 06:16:37 AM »

Question is whether CaCO3 is the only source of carbon dioxide. If it is - your reaction equation is all that you need to find the correct answer.

Note: don't round down intermediate results. I mean - write them here rounded down, but use full available precision when entering them into next steps of calculations. I got 1.53 g of CaCO3 and I bet our results differ (1.52g vs 1.53g) because you have rounded down number of moles to 0.0152.
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