Elemental S reacts with O2 to form SO3 according to the reaction:
2S+3O2→2SO3
How many O2 molecules are needed to react with 5.27 g of S?
Express your answer numerically in units of molecules.
=????molecules
Part B
What is the theoretical yield of SO3 produced by 5.27 g of S?
Express your answer numerically in grams.
= ???g
Subtitle: Limiting reactant
Next, consider a situation in which all of the S is consumed before all of the O2 reacts, or one in which you have excess S because all of the O2 has been used up.
For each of the given situations, indicate whether S or O2 is the limiting reactant.
A) 3.0 mol Sulfur & 5.0mol oxygen
B)3.0 mol sulfur & 4.0 mol oxygen
C) 3.0 mol sulfur & 3.0 mol oxygen
So for Part A it would be 1.48694^23?
Part B 13.21g?
I want to double check to get them right
Part C I was able to get them right, thank you so much.
Elemental S reacts with O2 to form SO3 according to the reaction:
2S+3O2→2SO3
How many O2 molecules are needed to react with 5.27 g of S?
Express your answer numerically in units of molecules.
=????molecules
mols S = g/atomic mass = 5.27/32 = 0.165
Convert to mols O2. That's 0.165 mols S x (3 mols O2/2 mols S) = 0.247
Convert to molecules O2 knowing that there are 6.02E23 molecules in 1 mol.
Part B
What is the theoretical yield of SO3 produced by 5.27 g of S?
Express your answer numerically in grams.
= ???g
From part A you know there are0.165 mols S. Looking at the equation, you get 1 mol SO3 for each mol S; therefore, you will get 0165 mol SO3. Convert to grams. grams = mols x molar mass = ?
Subtitle: Limiting reactant
Next, consider a situation in which all of the S is consumed before all of the O2 reacts, or one in which you have excess S because all of the O2 has been used up.
For each of the given situations, indicate whether S or O2 is the limiting reactant.
A) 3.0 mol Sulfur & 5.0mol oxygen
I do these the long way. Convert EACH to mols SO3 formed.
For S: 3.0 mols S x (2 mol SO3/2 mols S) = 3 mols SO3
For O: 5.0 mols O2 x (2 mols SO3/3 mols S) = 3.33 mols SO3.
You can form ONLY the smallest mols SO3; therefore, S is the limiting reagent (LR) and O2 will be the excess reagent (ER). The last two are done the same way.
Be sure to confirm all of the calculations AND make sure the numbers are rounded to the correct number of significant figures. I didn't check any of that.
B)3.0 mol sulfur & 4.0 mol oxygen
C) 3.0 mol sulfur & 3.0 mol oxygen
In the equation S(s) + O2 (g)-----} 2SO2 (g). How many oxygen atoms are in the product 2SO2 (g)?
A) Well, let's do the math. We have 3.0 mol of sulfur and 5.0 mol of oxygen. According to the balanced equation, 2 mol of sulfur react with 3 mol of oxygen. So, if we divide the number of moles of sulfur by 2 and the number of moles of oxygen by 3, we can determine the ratio between the two.
For sulfur: 3.0 mol sulfur / 2 = 1.5
For oxygen: 5.0 mol oxygen / 3 = 1.67
Since the ratio for sulfur is smaller than the ratio for oxygen, sulfur is the limiting reactant.
B) Now, we have 3.0 mol of sulfur and 4.0 mol of oxygen. Repeating the same process as before:
For sulfur: 3.0 mol sulfur / 2 = 1.5
For oxygen: 4.0 mol oxygen / 3 = 1.33
Once again, the ratio for sulfur is smaller than the ratio for oxygen, so sulfur is the limiting reactant.
C) Lastly, with 3.0 mol of sulfur and 3.0 mol of oxygen:
For sulfur: 3.0 mol sulfur / 2 = 1.5
For oxygen: 3.0 mol oxygen / 3 = 1.0
Surprisingly, both ratios are the same. In this case, neither sulfur nor oxygen is the limiting reactant.
Now, I hope you're enjoying this magical journey of determining limiting reactants! Remember, in this chaotic world of chemical reactions, everything is just a matter of ratios and proportions.
To determine the number of O2 molecules needed to react with 5.27 g of S, we need to use the balanced equation:
2S + 3O2 → 2SO3
First, we need to find the molar mass of Sulfur (S) from the periodic table. It is approximately 32.06 g/mol.
Next, we can convert the mass of Sulfur to moles using the following formula:
moles = mass / molar mass
moles of Sulfur = 5.27 g / 32.06 g/mol
moles of Sulfur ≈ 0.164 moles
According to the balanced equation, the ratio between Sulfur (S) and Oxygen (O2) is 2:3. This means that 2 moles of Sulfur react with 3 moles of Oxygen.
Now, using the mole ratio from the balanced equation, we can calculate the required moles of O2:
moles of O2 = (moles of Sulfur) * (3 moles of O2 / 2 moles of Sulfur)
moles of O2 = 0.164 moles * (3/2)
moles of O2 ≈ 0.246 moles
To convert from moles to molecules, we need to multiply by Avogadro's number, which is approximately 6.022 x 10^23 molecules/mol.
number of O2 molecules = (moles of O2) * (6.022 x 10^23 molecules/mol)
number of O2 molecules ≈ 0.246 moles * (6.022 x 10^23 molecules/mol)
number of O2 molecules ≈ 1.48152 x 10^23 molecules
Therefore, the number of O2 molecules needed to react with 5.27 g of S is approximately 1.48152 x 10^23 molecules.
Moving on to part B:
To find the theoretical yield of SO3 produced by 5.27 g of S, we need to use the stoichiometry of the balanced equation.
The molar mass of SO3 (Sulfur trioxide) can be calculated:
Sulfur (S) = 32.06 g/mol
Oxygen (O) = 16.00 g/mol (there are three of them in SO3)
Molar mass of SO3 = (32.06 g/mol) + (16.00 g/mol * 3) = 80.06 g/mol
Next, we can calculate the moles of SO3 produced using the following formula:
moles of SO3 = (mass of S) / (molar mass of S)
moles of SO3 = 5.27 g / 32.06 g/mol
moles of SO3 ≈ 0.164 moles
According to the balanced equation, the ratio between Sulfur (S) and Sulfur trioxide (SO3) is 2:2. This means that 2 moles of Sulfur react to form 2 moles of SO3. Therefore, the stoichiometry is 1:1 for S and SO3.
Since the stoichiometry is 1:1, the moles of SO3 produced (the theoretical yield) will be the same as the moles of S used.
Therefore, the theoretical yield of SO3 produced by 5.27 g of S is approximately 0.164 moles, which can be converted to grams by multiplying by the molar mass of SO3:
theoretical yield of SO3 = (moles of SO3) * (molar mass of SO3)
theoretical yield of SO3 ≈ 0.164 moles * 80.06 g/mol
theoretical yield of SO3 ≈ 13.124 g
Therefore, the theoretical yield of SO3 produced by 5.27 g of S is approximately 13.124 g.
In the next section, we need to determine if S or O2 is the limiting reactant in each given situation.
A) 3.0 mol Sulfur & 5.0 mol Oxygen:
According to the balanced equation, the ratio between Sulfur and Oxygen is 2:3. This means that for every 2 moles of Sulfur, we need 3 moles of Oxygen.
The mole ratio in the given situation is 3.0 mol Sulfur : 5.0 mol Oxygen, which is a 1.5:1 ratio. Since there is more Oxygen than the ratio requires, Oxygen is in excess and Sulfur is the limiting reactant.
B) 3.0 mol Sulfur & 4.0 mol Oxygen:
According to the balanced equation, the ratio between Sulfur and Oxygen is 2:3. The mole ratio in the given situation is 3.0 mol Sulfur : 4.0 mol Oxygen, which is a 0.75:1 ratio. As the ratio requires more Oxygen than what is available, Sulfur is in excess and Oxygen is the limiting reactant.
C) 3.0 mol Sulfur & 3.0 mol Oxygen:
According to the balanced equation, the ratio between Sulfur and Oxygen is 2:3. The mole ratio in the given situation is 3.0 mol Sulfur: 3.0 mol Oxygen, which is a 1:1 ratio. In this case, the ratio is exactly what is needed, so neither Sulfur nor Oxygen is in excess. Therefore, there is no limiting reactant; they are in perfect stoichiometric proportions.