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.
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?
5 answers
3. Using the coefficients in the balanced equation, convert moles Sb2S3 to moles Sb.
i don't get this step?
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.
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.
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.
1 answer