Workbook Self Test pg. 11 #8

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Diondraya_Taylor_1C
Posts: 23
Joined: Wed Sep 21, 2016 2:59 pm

Workbook Self Test pg. 11 #8

Postby Diondraya_Taylor_1C » Tue Sep 27, 2016 5:14 pm

Workbook Self Test pg. 11 #8

"8. How many moles of CO2 (g) is produced when 500g of CaCO3 (s) is used to neutralize an acid spill?
The equation for the reaction at 1 atm and 25 degrees C is:

CaCo3 (s) + H2SO4 (aq) --> CaSO4 (s) + CO2 (g) + H2O (l)"

In this sort of problem, when "neutralize" is used, does that mean that we know that the other reactant is the same sized sample? So here, would H2SO4 also be 500g making the reactant side 1000g and therefore the product side 1000g?

Kevin Bangi 1A
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Joined: Wed Sep 21, 2016 2:59 pm

Re: Workbook Self Test pg. 11 #8

Postby Kevin Bangi 1A » Tue Sep 27, 2016 5:37 pm

First off, the problem asks for moles of CO2 and not the grams of both formulas. So, we first look at the question, moles are the one needed to be found, to be exact, moles of CO2. In order to find the number of CO2 moles needed, Dimensional Analysis would be the method we use to convert from grams to moles.
1) Check if equation is balanced
2) Use the given value as the starting point
3) Dimensional Analysis
500gCaCO3 1 mole CaCO3 1 mole CO2
-------------- X ----------------- X ------------------
100.9g CaCO3 1 mole CaCO3
So the answer become 4.99 moles of CO2 is produced

Maggie Bui 1H
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Re: Workbook Self Test pg. 11 #8

Postby Maggie Bui 1H » Wed Sep 28, 2016 12:09 pm

It's not so much when the word "neutralize" is used, but if you're given the mass of only one reactant, I think it's safe to assume that the other reactant is present in excess. Therefore, it wouldn't affect the amount of product produced, so you can proceed with calculations using your given mass.


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