Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: zeus on April 22, 2004, 01:42:27 PM
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???
Hello I have the answer to the (a) but do not know what direction to go to obtain the answer to (b) or (c).
Zinc and sulfur react to form zinc sulfide according to the equation. Zn + S = ZnS
If 25.0g of ainc and 30.0g of sulfur are mixed,
(a) Which chemical is the limiting reactant?
25.0 Zn x 1molZn/63.39g Zn x1mol ZnS/1 mol Zn x 97.43 ZnS/1mol ZnS = 38.4 Zn
30.0g S x 1molS/32.07g S x 1mol ZnS/1mol S x 32.06 S/ 1mol ZnS = 30 S
(b) How many grams of ZnS will be formed?
UPDATE 4/23/2004 --answer: 97.43 ZnS Am I correct?
(c) How many ionic units of ZnS will be formed? Still do not know how to figure this out.
:'(
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For part A) cross out your units you will see that you didn't end with Zn but ZnS. For the second part of part A the last term should of been (MW of ZnS/ 1mol ZnS)
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You've gone a little far with your answer for part A... to find out which is the limiting reactant, you'll have to stop once you get to mols of Zn and mols of S. You did the right thing in finding out the mols of Zn and S, so let's go from there.
Your calculations: (FW = Formula Weight)
25.0g Zn / 63.39g/mol Zn FW = 0.39mol Zn
30.0g S / 32.07g/mol S FW = 0.93mol S
Now, take a look at your equation:
Zn + S = ZnS
The stoichiometry dictates that for every mol of Zn you use, you use one mol of S. Remember, mols = mols, but grams don't equal grams.
So, if you have 0.39mol Zn and 0.93mol of S, how many mols of ZnS can you make, total? Only 0.39... the extra 0.54mol of S just sits there and does nothing because it can't react. (More correctly, it may melt, because this reaction is very exothermic, but melting isn't reacting). You now know which is the limiting reactant.
Knowing that only 0.39mol of ZnS can be formed, assuming there are no problems while performing the reaction, you can calculate the number of grams of ZnS formed by using the formula weight. If you have questions about this, ask.