If 3.68 grams of zinc were allowed to react with excess hydrochloric acid to produce zinc chloride and hydrogen gas, how much zinc chloride should be produced? Suppose that 7.12 grams of zinc chloride is actually recovered, what is the per cent yield?

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Here... Since there is excess HCl, you know the Zinc will be the limiting reagent.  I was able to create the reaction equation based on the info you gave: Z + 2HCl > ZCl2 +2H.  Now, in order to find the theoretical yield of ZCl, you need to convert Zinc from grams to moles.  Do this by dividing by its molar mass (65.38), thus: 3.68 (g) / 65.38 (g/mol) = 0.056286... moles.  

Now, using the balanced equation we made earlier, we see that 1 mole of Zinc creates 1 mole of ZCl2.  Thus our ratio is 1:1.  This makes the next step easy.  Since it is 1:1, we multiply the number of moles we have of Zinc (0.056286...) by the number of moles it will create of ZCl (1).  (0.056286)(1) = 0.056286 moles ZCl.  Now convert this to grams by multiplying by its molar mass (136.28) and you get 7.67 grams.  This is your theoretical yield.  The percent yield is found by dividing the actual amount obtained (7.12 g.) by the theoretical yield (7.67 g.) then multiply that by 100%.   When this is done, your Percent Yield is about 92.8%

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