Posted by sam on Friday, May 11, 2007 at 1:52pm.
if 20.2g of zn react with and equal amount of magnesium chloride to form magnesium and zinc chloride[zn+mgcl2>
a.what is the limiting reactant?
b. how much magnesium will be produced?
Here is the way to solve most of these stoichiometry problems.
Step 1. Write the balanced equation.
Zn + MgCl2 ==> Mg + ZnCl2
Step 2. Convert what you have into mols.
2a. mols Zn = 20.2/65.4 = 0.309
2b. mols MgCl2 = 20.2/95.2 = 0.212
Step 3. Using the coefficients in the balanced equation, convert each of the mols from what you have (in this case mols Zn and mols MgCl2) to mols of what you want (in this case Mg is the product).
3a. First from 2a (mols Zn).
mols Mg = mols Zn x (1 mol Mg/1 mol Zn) = 0.309 x 1/1 = 0.309 mols Mg product.
3b. from 2b (mols MgCl2).
mols Mg = mols MgCl2 x (1 mol Mg/1 mol MgCl2) = 0.212 x 1/1 = 0.212.
3c. Now you make a decision. Both answers can't be correct; i.e., 3a says 0.309 mol Mg will be produced while 3b says 0.212 mols Mg will be produced. The answer ALWAYS is the smaller one AND MgCl2, then, is the limiting reagent. (That answers part a of the problem.)
Step 4. The problem doesn't state if you want mols, grams, or what. If you want mols, then 0.212 mol Mg will be produced. If you want grams, then
mols Mg x atomic mass Mg = grams Mg.
If you haven't seen this method before, it might be good to learn it. It will work literally hundreds of problems for you. If you don't have a limiting reagent problem, and many are not, simply omit steps 2b, 3b, and 3c.
You need to go through and correct any numbers for atomic masses of Zn, Mg, and molar mass MgCl2 for I estimated those values. They are close but may not be the best. Look them up on the periodic table.
Finally, I want you to know that this reaction will not take place but the reverse one will. That is, Mg will displace Zn from ZnCl2 in this reaction
Mg + ZnCl2 ==> Zn + MgCl2 but the reaction will not take place as it is writtenin the problem. Technically, then, the answer to the problem is that NO Mg will be produced; however, I don't think that is the intent of the problem. The steps I showed are the actual steps you will need for a valid reaction and they work equally well for reactions that will not occur.
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