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Home > Science > Chemistry   »   [HELP]gr11 chemistry - predicting chemical reactions and balancing chemical equations

 
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Old Nov 27, 2004, 01:05 PM
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[HELP]gr11 chemistry - predicting chemical reactions and balancing chemical equations

hey can nebody help me with my chemistry hmwk? i dun really get it... thanks!

For each of the following reactions, determine what the products of each reaction will be. When you have predicted the products, balance the equation and use a table of solubility products to determine which of the products (if any) will precipitate. Assume all reactions take place in water.

3) NaC2H3O2 + H2SO4 -->





4) Cu(OH)2 + H3PO4 -->





6) Zn + H2CO3 -->







For each of the following questions, classify the reaction type (synthesis, decomposition, combustion, single replacement, double replacement, or other) and predict the balanced chemical equation. Provide a word equation as well.



6. Sulfuric acid is neutralized by aqueous sodium hydroxide.






9. Propane burns in air.






10. Na2CO3(aq) + HCl(aq) --> NaCl(aq) + CO2(g) + H2O(l)







Indicate which of the following double displacement reactions will go to completion, and circle what the product of the reaction will be (if any). You may want to consult a list of Ksp values to solve these problems.



1)NaOH + CaBr2
-->



Will this reaction occur? ____



2)Pb(NO3)2 + HCl -->



Will this reaction occur? ____



3)Na2CO3 + KF
-->



Will this reaction occur? ____



4)AgNO3 + CuSO4
-->



Will this reaction occur? ____



5)AgF + NiCl2
-->



Will this reaction occur? ____

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Old Nov 27, 2004, 06:00 PM   #2  
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reaction type

Reaction type is pretty straight forward, there are key things to look for.
Anything dealing with propane or a gas will most likely be a combustion reaction. Another way to know is that a combustion reaction will result in the production of C20 and H2O.
Decompositon is when you have a compound produces two more substances.
Oka, Here's a simple way to remember:

Formation(synthesis/composition)
E + E --> C
element + element --> compound
ie. 2Na + Cl2 --> 2NaCl
So you'll have a product which is a combination of the two reactants.

Decomposition Reaction
C --> E + E
Pretty much a reverse of the above.
ie. 2HgO --> 2Hg + O2

Combustion (think of 'burn')
Hydrocarbon + O2(g) --> CO2(g) + H2O(g)
The O2(g) --> CO2(g) + H2O(g) will always remain the same, never change, in these reactions.

Single and Double Replacement are really simple and straight forward.
Single Replacement
E + C --> new E + new C
so ie.
Cu(s) + ZnO(s) --> CuO(s) + Zn(s)
You swap the Cu and the Zn, you do this because they are both 'positive'

Double Replacement
C + C --> newC + newC
ie. NaCl(s) + KBr(s) --> NaBr + KCl
You can swap either the positives or the negatives that are before the arrow.
In this case the two positives are swapped again. Na and K to form the two new compounds.

So when you look at a question, write out what you know, figure out what you need to know and what needs to be formed.

Na2CO3(aq) + HCl(aq) --> NaCl(aq) + CO2(g) + H2O(l)
aqueous sodium carbonate + aqueous hydrochloric acid --> sodium chloride + carbon dioxide + water
You have compound+compound --> compound + element + element
Think about it........

Once you learn the types of reactions everything will come easier.
Like you asked what about
Zn + H2CO3 -->
you have an element + compound
so you're going to have a single replacement reaction occuring, right?
You need a new element and a new compound to be formed.
Think you can figure it out?

I know this isn't much help but I hope it gave you a start. If you're confused about anything let me know.
This is a great site to help you out, so check it out if in doubt okay.
http://www.chemtutor.com/react.htm

Good luck.
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Old Nov 28, 2004, 07:04 PM   #3  
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thanks for the site..i will look over it

how do i find a tutor? therez no results..lol
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Old Nov 29, 2004, 09:12 AM   #4  
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tutor

You want to find a tutor?
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Old Nov 30, 2004, 02:23 PM   #5  
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yea..i wanna find one on that site...but that's ok..and thx for ur help
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Old Nov 30, 2004, 03:39 PM   #6  
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okay

Alright, good luck. There are some good sites with tutors, but I don't know the names right now.
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