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Topic: Balancing  (Read 29453 times)

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Offline Shea

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Balancing
« on: July 31, 2006, 02:41:03 PM »
Could someone teach me how to write the balanced equation for this question?

How many g of silver chloride will be produced by reacting 10g of silver nitrate with sodium chloride?
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Offline Borek

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Re: Balancing
« Reply #1 on: July 31, 2006, 02:44:45 PM »
Write formulas of all substances mentioned. Which are reactants and should be present on the left? Which are products and should be present on the right? Is there something left?

http://www.chembuddy.com/?left=balancing-stoichiometry&right=balancing-reactions
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Offline Shea

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Re: Balancing
« Reply #2 on: July 31, 2006, 02:48:02 PM »
What, do I just write, "AgNO3 + NaCl -> AgCl ??"
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Offline Borek

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Re: Balancing
« Reply #3 on: July 31, 2006, 02:58:10 PM »
Almost OK - but check that some elements are not present on the right (like Na).
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Offline Shea

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Re: Balancing
« Reply #4 on: July 31, 2006, 04:00:51 PM »
How does it appear on the right side?  Do I just put Na over there?
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Offline Borek

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Re: Balancing
« Reply #5 on: July 31, 2006, 04:13:44 PM »
Look at ions present - there is one other (complex) ion that is not present on the right.
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Offline sdekivit

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Re: Balancing
« Reply #6 on: July 31, 2006, 04:16:26 PM »
You add AgNO3 to NaCl. In solution they will by hydrated and the following ions are formed:

Ag+, NO3-, Na+, Cl-

Since AgCl is formed, what ions will form the second salt in this displacement reaction ?
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Offline Shea

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Re: Balancing
« Reply #7 on: July 31, 2006, 04:27:00 PM »
I'm a chemistry noob, I know, but tell me, how am I supposed to know that they'll be hydrated, and what the ions are?

Is it NaNO3?
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Offline sdekivit

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Re: Balancing
« Reply #8 on: July 31, 2006, 04:30:52 PM »
yes it's NaNO3 :)

About the ions: learn them :) it's the only way.
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Offline Shea

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Re: Balancing
« Reply #9 on: July 31, 2006, 04:36:37 PM »
Well, I just guessed that. 

You said it was a salt so I looked up something on wikipedia and I saw NaNO3...

Can you tell me how knowing the ions will help me learn to write these equations?  And where can I find a list or something of these ions?
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Offline Borek

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Re: Balancing
« Reply #10 on: July 31, 2006, 04:41:25 PM »
Quote from: Shea on July 31, 2006, 04:27:00 PM
tell me, how am I supposed to know that they'll be hydrated

All ions in water solutions are hydrated. And you are supposed to know such things - you are studying chemistry, aren't you? ;) But it is not relevant for this question.

Quote from: Shea on July 31, 2006, 04:36:37 PM
Can you tell me how knowing the ions will help me learn to write these equations?  And where can I find a list or something of these ions?

I don't think there is such thing as a list of ions - but you should be probably able to find simple rules that describe how to find their charges.

How they will help you? You know AgCl precipitated from the solution, you just remove these preciptated ions (Ag+, Cl-) from the list - and you are left with other ions, they form a salt - NaNO3.
« Last Edit: July 31, 2006, 04:47:31 PM by Borek »
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Offline swati

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Re: Balancing
« Reply #11 on: July 31, 2006, 04:47:53 PM »
:)  So the equation becomes
    AgNO3   +    NaCl    ----->     AgCl     +    NaNO3

Quote from: Shea on July 31, 2006, 02:41:03 PM
How many g of silver chloride will be produced by reacting 10g of silver nitrate with sodium chloride?
Since 1 mole of AgNO3  gives 1mole of AgCl on reaction with NaCl, so calculate how much moles are there in 10 g of AgNO3 . Thus you can easily find the number of moles of AgCl formed which can then be converted to grams.
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Offline Shea

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Re: Balancing
« Reply #12 on: July 31, 2006, 04:57:29 PM »
And I do that by finding the molar weight of AgNO3, then dividing that by 10, and that will be how many mols there are?

How do I go from mols to grams?  Do I multiply mols by the molar weight?
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Offline swati

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Re: Balancing
« Reply #13 on: July 31, 2006, 05:20:27 PM »
Molar weight of AgNO3 = 170 g

170 g of AgNO3 = 1 mole of AgNO3
 10 g of AgNO3  = 10 / 170 mole of AgNO3
                            = 0.059 mole of AgNO3

So , 0.059  mole of AgNO3 are present .

0.059 mole of AgNO3  -------->  0.059 mole of AgCl
1        mole of AgCl = 143.5 g of AgCl
0.059 mole of AgCl  =143.5 * 0.059 g of AgCl = 8.467 g of Ag Cl

Quote
In short
Number of moles = Mass of the compound / Molar Mass

Mass of the compound = Number of moles * Molar Mass
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Offline Shea

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Re: Balancing
« Reply #14 on: July 31, 2006, 05:22:17 PM »
Thats exactly what I needed to see.  Thanks. 
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