if you react 8.60g of Sb with 11.4g of I and recover 10.9 of SbI, what is the percentage yield of antimony iodide
2Sb + 3I2 -> 2SbI3
Note your typo at the beginning. You obtain 10.9 g SbI3, This is a limiting reagent (LR) problem. I do these the long way.
2Sb + 3I2 -> 2SbI3
mols Sb = grams/atomic mass Sb = 8.60/121.8 = estimated 0.07
mols I = 11,4/126.9 = estimated 0.09mols SI3 formed
mols SbI3 formed with Sb and excess I2
0.07 x (2 mols SbI3/2 mols Sb) = 0.07
mols SbI3 formed from 0.09 mols I2 and excess Sb,
0.09 mols I2 x (2 mols SbI3/3 mols I2) = 0.06
Since I2 will form fewer mols SbI3 then I2 is the LR and Sb is the excess reagent (ER)
grams SbI3 formed = mols SbI3 x molar mass SbI3 = ? This is the theoretical yield (TY) which assumes it is 100% efficient.
The actual yield (AY) from the problem is 10.9 grams.
% yield = (AY/TY)*100 = ?
Note that I've estimated the math calculations so you should repeat all of them. Post your work if you get stuck. You might want to confirm the atomic masses to make sure I looked them up properly.
Well, it looks like you're dealing with a chemical reaction here. But let me put it in simpler terms. If you mix 8.60 grams of Sb with 11.4 grams of I and end up with 10.9 grams of SbI, how much of a joke is the percentage yield? Well, let's calculate it!
To find the percentage yield, you need to divide the actual yield (10.9g) by the theoretical yield (the maximum amount that could have been produced if everything went perfectly).
Now, let's take a closer look at the balanced equation: 2Sb + 3I2 -> 2SbI3. From this equation, you can see that the molar ratio between Sb and SbI is 2 to 2, which means that for every 2 moles of Sb, you should theoretically get 2 moles of SbI.
To find the theoretical yield of SbI, you'll need to convert both Sb and I to moles. The molar mass of Sb is 121.8 g/mol, so 8.60g of Sb is equal to 0.0707 moles. Similarly, the molar mass of I is 253.8 g/mol, so 11.4 g of I is equal to 0.0449 moles.
Since the molar ratio is 1:1 between Sb and SbI, the number of moles of SbI that could be produced is also 0.0707 moles.
But wait, we're given that the actual yield is 10.9 grams of SbI, which is equal to 0.0276 moles.
To find the percentage yield, divide the actual yield (0.0276 moles) by the theoretical yield (0.0707 moles) and multiply by 100.
(0.0276/0.0707) x 100 = 39.1%
So, the percentage yield of antimony iodide in this reaction is approximately 39.1%. I guess you could say it's like a clown's hat - it's not quite perfect, but it still puts on a good show!
To calculate the percentage yield of antimony iodide (SbI3), we need to compare the actual yield to the theoretical yield.
1. Calculate the molar mass of antimony (Sb) and iodine (I):
- Sb: 121.76 g/mol
- I: 126.90 g/mol
2. Determine the limiting reactant:
The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed. To find the limiting reactant, we need to compare the amount of each reactant to the stoichiometry of the balanced equation.
- Sb: 8.60 g
- I: 11.4 g
Convert the masses to moles using the molar mass:
- Moles of Sb: 8.60 g / 121.76 g/mol = 0.0706 mol
- Moles of I: 11.4 g / 126.90 g/mol = 0.0898 mol
Now compare the moles to the stoichiometry of the balanced equation (2Sb:3I2:2SbI3):
Sb: 0.0706 mol
I: 0.0898 mol / 3 (since 3 moles of I2 are required to react with 2 moles of Sb) = 0.0299 mol
The limiting reactant is iodine (I) since it produces fewer moles of SbI3 than Sb.
3. Calculate the theoretical yield of SbI3:
Using the stoichiometry of the balanced equation, we can determine the theoretical yield of SbI3 when Sb is the limiting reactant.
From the balanced equation: 2 moles of Sb react to form 2 moles of SbI3.
Since 0.0299 mol of I is required to react, the theoretical yield of SbI3 is:
(0.0299 mol of I) × (2 mol of SbI3 / 3 mol of I) × (331.60 g/mol of SbI3)
≈ 3.54 g
4. Calculate the percentage yield:
Percentage yield = (actual yield / theoretical yield) × 100%
Given: actual yield = 10.9 g, theoretical yield = 3.54 g
Percentage yield = (10.9 g / 3.54 g) × 100% ≈ 307.3%
Therefore, the percentage yield of antimony iodide is approximately 307.3%.
To determine the percentage yield of antimony iodide (SbI3), you need to compare the actual yield of SbI3 obtained (10.9g) with the theoretical yield that could be obtained based on the amounts of reactants used.
First, let's calculate the molar masses of the substances involved:
- Molar mass of Sb (antimony) = 121.76 g/mol
- Molar mass of I (iodine) = 126.90 g/mol
- Molar mass of SbI3 (antimony iodide) = (121.76 g/mol) + (3 * 126.90 g/mol) = 500.36 g/mol
Next, let's determine the moles of Sb and I used in the reaction:
- Moles of Sb = mass of Sb / molar mass of Sb
= 8.60g / 121.76 g/mol = 0.0706 mol
- Moles of I = mass of I / molar mass of I
= 11.4g / 126.90 g/mol = 0.0897 mol
According to the balanced chemical equation, the stoichiometry of the reaction is 2:3. This means that 2 moles of Sb react with 3 moles of I to form 2 moles of SbI3.
From the stoichiometry, we can calculate the theoretical yield of SbI3:
- Moles of SbI3 formed = (moles of Sb) / 2
= 0.0706 mol / 2 = 0.0353 mol
Now, let's calculate the theoretical mass of SbI3:
- Mass of SbI3 = (moles of SbI3 formed) * (molar mass of SbI3)
= 0.0353 mol * 500.36 g/mol = 17.67g
Finally, we can calculate the percentage yield:
- Percentage yield = (actual yield / theoretical yield) * 100
= (10.9g / 17.67g) * 100 = 61.8%
Therefore, the percentage yield of antimony iodide (SbI3) in this reaction is approximately 61.8%.