A sample of 4.5g aluminum reacts with 6.8g of sulfur to form aluminum sulfide according to the following reaction.

2Al+3S----->Al2S3

Which one is the limiting reagent?
Al is limiting reagent.
Because
Al=(4.5g)(1mol/26.98g)(1mol/2mol)(150.17g Al2S3/1mol Al2S3)=1.25g
S=(6.8g)(1mol/96.21g)(1molAl2S3/3mol S)(150.17g Al2S3/1mol)=3.54g
So, limiting reagent is Al.

So, my question is

1.what is the mass of aluminum sulfide formed?
I thinks its 1.25g, If it is not, please teach me how to figure out

2.what is the theoretical yield?
I think its 1.25g, If it is not, please teach me how to figure out

3.which reagent is in excess?
I think its S. If it is not, please teach me how to figure out

4.what is the mass of excess reagent?
(1.25g Al2S3)(1moeAl2S3/150.17g)(1molS/1molAl2S3)(32.07gS/3molS)=0.08898g
Im really not sure about this question , so please check this calculation for me. Thanks.

5.if 8.2g of product was formed, what is percent yield?
(8.2/1.25)x100=656
I am really not sure about this one cause I think that's too much percentage.. weird.

Please Please help me Thanks

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.





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.

1.what is the mass of aluminum sulfide formed?
So, its just adding 4.5g and 6.8g? So its 11.3g??

Remember that what determines whether a reagent is limiting or not is the amount of moles, not the mass.

1. You know that there are 0.21 moles of sulfur and 0.17 moles of aluminium.
You also know that only 0.21 moles of sulfur react. The mole ratio of sulfur to the product is 3:1 (3 moles of sulfur gives one mole of Al2S3). Find the number of moles of Al2S3 produced, and then its mass. (I did not get 1.25 g, but a much higher mass)

So mass of aluminum sulfide formed is 29.4g?

Nope, way to far this time..

You have 0.21 moles of sulfur.
You obtain a third of that amount of moles for aluminium sulfide, so you get 0.07 moles of Al2S3.

0.07 moles of Al2S3 has a mass of (0.07*150) = 10.6 g (taking the moles as 0.2125 mol, initially)

Yes I got that number thanks

Could please help me how to find this one?

4.what is the mass of excess reagent?

Now that you know Al is in excess, find the amount of moles which do not react.

That makes 0.167 - 0.14 = 0.0267 moles

Can you find that mass of aluminium now?