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posted by  MzKlassy on 10/27/2008 4:23:45 PM  |  status: Closed  

PLEASE HELP ME FIGURE THE MASS

Course Textbook Chapter Problem
General Chemistry N/A N/A N/A
Question Details:
You carefully weigh out 15.00 g of \rm CaCO_3 powder and add it to 60.75 g of \rm HCl solution. You notice bubbles as a reaction takes place. You then weigh the resulting solution and find that it has a mass of 69.60 g. The relevant equation is

\rm CaCO_3{\it (s)}+2HCl{\it (aq)}\rightarrow H_2O{\it (l)}+CO_2{\it (g)}+CaCl_2{\it (aq)}

What mass of \rm CO_2 was produced in this reaction?

Assume that the reaction goes to 100% completion producing only the three desired products.
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posted by Arcanum on 10/27/2008 5:22:04 PM  |  status: Live
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MzKlassy's comment:
"Thank you so much for explaining this as well I understand it a lot better now Thank You Again."
Response Details:
Ok, this problem looks huge, which may scare a lot of people but once you think about it, its not all that scary. Lets go through the idea that mass can not be created nor destroyed, the Conservation of mass. Simple way, is that you add the two masses you have together to get your total mass...
 
15.00g CaCO3 + 60.75g HCl = 75.75grams total thats what was put into the equation as a whole.
 
Now, we had a final measurement of 69.60grams which is what was measured, thus we just need to do
 
75.75g - 69.60g = 6.15grams of CO2
 
Now you may ask how? Well if you look at the equation, CO2 is a gas, and it can not be measured like a solid or a liquid, there is a diffrent way in measuring it. The 69.60grams is the measurement of the water and the CaCl2 the measurement did not say it was just the CaCl2 so water is a factor in that, which makes up that weight. And, if this is a closed system, where nothing can get in or out,then just remember the conservation of mass and you will do great :+)
 
I hope this helps you out a little bit, if you need anymore help with this, please PM and ill try my best :+)
Pffhh, if I knew everything, I would spend my entire time on here answering questions, but until then ill spend my entire time on here asking questions... :+)
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