Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Lindsay on November 11, 2008, 10:22:22 PM
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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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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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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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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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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.