Heating an ore of antimony (Sb2S3) in the presence of iron gives the element antimony and iron(II) sulfide.
Sb2S3(s) + 3 Fe(s) 2 Sb(s) + 3 FeS(s)
When 10.9 g Sb2S3 reacts with an excess of Fe, 9.84 g Sb is produced. What is the percent yield of this reaction?
You don't have an arrow in the equation. I assume you know where it goes. Place it there, first thing.
2. Convert 10.9 g Sb2S3 to moles. moles = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles Sb2S3 to moles Sb.
4. Now convert moles Sb to grams. grams = moles x molar mass. This is the theoretical yield.
5. %yield = (actual yield/theoretical yield) * 100 = ??
actual yield is given in the problem of 9.84 g.
3. Using the coefficients in the balanced equation, convert moles Sb2S3 to moles Sb.
i don't get this step?
Sb2S3(s) + 3 Fe(s) 2 Sb(s) + 3 FeS(s)
xx moles Sb2S3 x (2 moles Sb/1 mole Sb2S3) = yy moles Sb.
The coefficient of Sb is 2 and that of Sb2S3 is 1. Note that the factor in parentheses converts moles Sb2S3 to moles Sb. How? Note that the unit Sb2Sb cancels but moles Sb are left and that's what you are looking for.
how many moles Sbs are left!?
What do you mean how many are left? All of them are left. You had none to start with. You now have xx moles. None are used up are they?
Sb2S3(s) + 3 Fe(s) 2 Sb(s) + 3 FeS(s)
When 10.9 g Sb2S3 reacts with an excess of Fe, 9.84 g Sb is produced. What is the percent yield of this reaction?
10.9/339.712 = 0.032 moles Sb2S3
0.032 mols Sb2S3 x (2 mols Sb/1 mol Sb2S3) = 0.032 x (2/1) = 0.064 mols Sb.
g Sb = moles Sb x molar mass Sb = 0.064 x 121.7 = 7.8 g
%yield = ???
I may have figured out your (our) problem. The theoretical yield I have is about 7.8 g (you need to go through and redo all of the calculations because I've estimated and rounded here and there) however, the problem tells us that 9.84 g Sb were produced. That can't be if we started with 10.9 g because 7.8 g is the theoretical yield. This would make the percent yield over 100% which isn't possible. Check your numbers to make sure they are right. Check my calculations and see if you see anything wrong. I've gone through it a couple of times and I don't see anything out of line.
To determine the percent yield of a reaction, we need to compare the actual yield (the amount of product obtained in the reaction) to the theoretical yield (the amount of product that would be obtained if the reaction went to completion).
First, we need to calculate the theoretical yield of Sb in the reaction.
From the balanced equation, we can see that the stoichiometric ratio between Sb2S3 and Sb is 1:2. This means that for every mole of Sb2S3 that reacts, we should obtain 2 moles of Sb.
To convert the given mass of Sb2S3 to moles, we need to know the molar mass of Sb2S3.
The molar mass of Sb is 121.76 g/mol, and the molar mass of S is 32.06 g/mol. So, the molar mass of Sb2S3 can be calculated as:
(2 * 121.76 g/mol) + (3 * 32.06 g/mol) = 339.68 g/mol
Now, we can calculate the number of moles of Sb2S3:
moles of Sb2S3 = mass / molar mass
moles of Sb2S3 = 10.9 g / 339.68 g/mol ≈ 0.0321 mol
According to the stoichiometry of the balanced equation, 0.0321 mol of Sb2S3 will produce 2 * 0.0321 mol of Sb. So, the theoretical yield of Sb is:
theoretical yield of Sb = (2 * 0.0321 mol) * (121.76 g/mol) ≈ 7.82 g
Now, we can calculate the percent yield:
percent yield = (actual yield / theoretical yield) * 100
percent yield = (9.84 g / 7.82 g) * 100 ≈ 125.57%
Therefore, the percent yield of this reaction is approximately 125.57%.