A sample of 4.5g aluminum reacts with 6.8g of sulfur to form aluminum sulfide according to the following reaction.
which one is the limiting reagent?
Al is limiting reagent.
 \frac {1mol}{26.98g} = 0.1668\,moles)
 \frac {1mol}{32.065g}=0.2121\,moles)
You need 2 moles of Al to react with 3 moles of S So, you have
Where one equivalent of Al would react with one equivalent of S. You have fewer equivalents of S so S is the limiting reagent.
1.what is the mass of aluminum sulfide formed?
The limiting reagent is S. You have 0.2121 moles of S. You will form 1 mole of Al2S3 from 3 moles of S. Therefore, you will form 0.0707 moles of Al2S3. You can figure out the weight (hint to check: You started with 4.5 g of Al; you started with 6.8 g of S. All of the S reacts; most of the Al reacts. You'll have over 10 g of product).
2. What is the theoretical yield?
The theoretical yield is what you calculated in 1, above.
3.which reagent is in excess?
You now know that it is Al.
4.what is the mass of excess reagent?
Subtract the number of equivalents of Al that will react from the equivalents of Al that you started with. Subtract the number of equivalents that are required in the reaction. Multiply the excess equivalents of Al by 2 to get the number of moles of Al. Convert that to the mass of Al.
5.if 8.2g of product was formed, what is percent yield?
8.2 / theoretical yield x 100 = percent yield.
Linear Mode
