Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Shea on July 31, 2006, 02:41:03 PM
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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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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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What, do I just write, "AgNO3 + NaCl -> AgCl ??"
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Almost OK - but check that some elements are not present on the right (like Na).
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How does it appear on the right side? Do I just put Na over there?
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Look at ions present - there is one other (complex) ion that is not present on the right.
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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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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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yes it's NaNO3 :)
About the ions: learn them :) it's the only way.
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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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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.
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.
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:) So the equation becomes
AgNO3 + NaCl -----> AgCl + NaNO3
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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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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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
In short
Number of moles = Mass of the compound / Molar Mass
Mass of the compound = Number of moles * Molar Mass
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Thats exactly what I needed to see. Thanks.
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Thats exactly what I needed to see. Thanks.
Welcome .
But never expect spoon feeding . If someone gave you this answer in the starting then you won't use your brain . Now you understood the concept & you won't forget it easily . :D
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How many g of zinc are required to produced 50g of hydrogen when reacting zinc with sulphuric acid?
Did I do this right --> Zn + H2SO4 -> 2H + ZnSO4
2 moles of H make 1 mole of Zn.
2g of H = 2 moles of H
50 g = 50 moles.
50 moles of H = 25 moles of Zn
25 moles of Zn = 1635 g.
1635 g of Zn * 25 = 40875 g
Did I mess up anywhere?
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Did I mess up anywhere?
Yes, twice, but these are minor mistakes, you understand how to do the question.
Did I do this right --> Zn + H2SO4 -> 2H + ZnSO4
H2, not 2H. Luckily, mass is the same.
25 moles of Zn = 1635 g.
1635 g of Zn * 25 = 40875 g
You have already calculated mass of 25 moles of Zn above, second multiplication is a mistake.
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So 1635 g of Zinc are required to produce 50g of hydrogen when reacting zinc with sulphuric acid?
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Yes. 1622g if you want to be really precise - or, 1.6*103g to be correct when it comes to significant digits.
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Ok, I guess that sorta counts as me doing one for myself for once.
Can you check my work on finding out how much sulphuric acid was needed for the previous question?
Zn + H2SO4 -> H2 + ZnSO4
2 moles of H = 1 mole of H2SO4
25 moles of H2SO4 = 2451.9625 g?
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Can you check my work on finding out how much sulphuric acid was needed for the previous question?
Zn + H2SO4 -> H2 + ZnSO4
2 moles of H = 1 mole of H2SO4
25 moles of H2SO4 = 2451.9625 g?
It is correct , but instead of 2 moles of H write 1 mole of H2 .
Note : Mass of 1 mole of H2 is also 2 g
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You may check it by yourself - just download and install trial version of EBAS (see link in my signature), start the program, mark the reaction equation as written in your post and copy it to clipboard, press Ctrl-F to create reaction from clipboard content, and enter 50 as hydrogen mass.
Your result looks more or less correct, although it abuses significant digits. EBAS shows 2433 g (in output frame, below H2SO4 formula), but it uses much more accurate molar masses. To be really correct you should give answer as 2.4*103 g.
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I've been trying to use that, and, for the most part, I respect it a lot. But what can it do for me if I can't even enter the proper equation?
Anyway, my teacher wants me to show work, so I have to do it myself.
I'm kinda getting the hang of it though.
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what can it do for me if I can't even enter the proper equation?
Nothing.
Anyway, my teacher wants me to show work, so I have to do it myself.
And I am not proposing you to rely solely on the program, just to check the results. I won't help you for the next 8 hours, it is after midnight here and I am changing position to more horizontal :)
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There is a partial list of ions on this site:
http://chemtutor.com/compoun.htm
also some text books have lists of ions. For example, Foundations of Chemistry 2nd editions by Toon and Ellis has a list of common ions.
Good luck :)
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Thanks for the list.
Now, this question seems to be different from the ones yesterday. How do I do it?
How many grams of fluorine are required to react with 100g of NaBr in the following reaction:
F2 + 2NaBr -> 2NaF + Br2
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It is identical. Try to use ratios, as described on my page:
http://www.chembuddy.com/?left=balancing-stoichiometry&right=ratio-proportions
or go for NaBr moles, F2 moles - and convert them back to mass.
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Do I have to do anything with the 2NaF?
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Do I have to do anything with the 2NaF?
What for? You are asked only about reactants, amounts of products doesn't matter.
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Well, the other questions involved a product. This one's different.
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Well, the other questions involved a product. This one's different.
It doesn't matter. You always start with reaction equation and molar ratios it defines.
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Well, how do I find the mass of F2 when the products don't matter and I only know that it's supposed to react with 100g of NaBr?
The program says the answer is 18.46g. What steps did it take to get that answer?
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The program says the answer is 18.46g.
At least you know how to correctly use EBAS ;)
100g NaBr - how many moles is it?
Look at the reaction equation. How many moles of F2 react with 1 mole of NaBr? How many moles F2 will react with known number of moles of NaBr?
It is identical with the calculations you did yesterday.
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EBAS says 100g of NaBr is .97 moles.
That means the equation says 1 mole of F2 and .97 moles of NaBr will make 2NaF + Br2?
If its just 1 mole of F2, than thats just 38g?
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EBAS says 100g of NaBr is .97 moles.
I hope you know how to calculate it by yourself.
That means the equation says 1 mole of F2 and .97 moles of NaBr will make 2NaF + Br2?
No. Look at the reaction equation and coefficients there:
F2 + 2NaBr -> 2NaF + Br2
It reads: 1 mole of F2 reacts with 2 moles of NaBr yielding 2 moles of NaF and 1 mole of Br2. Now the question is: if 2 moles of NaBr react with 1 mole of F2, how many moles of F2 will react with 0.97 mole of NaBr?
It is described in the lecture I have pointed you to earlier tonight.
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Yeah, I can calculate that myself. I can do all the calculations, the hard part is just knowing what to calculate.
But, thanks to your rephrasing of the question, I'm not as confused anymore.
That's why EBAS says .4859 moles for F2.
Thats just a simple ratio.
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Thats just a simple ratio.
Told you so :) Once you understand that almost all stoichiometric questions are identical.