I don't understand this question?
What is the maximum amount in grams of SO3 that can be produced by the reaction 1.0 g of S with 1.0 g of O2
EQUATION: 2 S(s) +3 O2(g)--->2 SO3(g)
The choices are:
A) 3.8
B) 2.5
C) 0.27
D) 1.7
Shouldn't it be 2.0, or am I forgetting to do something
Thanks!
Well, well, well, it's chemistry time! Don't worry, my friend, I've got your back.
First, we need to figure out which reactant is limiting the production of SO3. To do that, we have to calculate the number of moles for both S and O2:
For S: 1.0 g / molar mass of S
For O2: 1.0 g / molar mass of O2
Now, let's compare the moles of S and O2. Since the reaction is in a 2:3 ratio, we can determine that the limiting reactant is the one with fewer moles.
Once we know the limiting reactant, we can use stoichiometry to find the maximum amount of SO3 produced. Keep in mind that the stoichiometric ratio is 2:2 (meaning 2 moles of S produce 2 moles of SO3).
Finally, using the molar mass of SO3, we can convert moles to grams to find the maximum amount of SO3 produced.
Now, I won't spoil the answer for you, but trust me, it's not exactly 2.0 grams. Take a closer look at those calculations, and I'm sure you'll find the right answer among the choices. Good luck, my chemistry buddy!
To find the maximum amount of SO3 that can be produced by the reaction, you need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
To do this, we need to find the number of moles of each reactant.
1. Calculate the number of moles of S:
molar mass of S = 32.06 g/mol
moles of S = mass of S / molar mass of S
moles of S = 1.0 g / 32.06 g/mol
moles of S ≈ 0.0312 mol
2. Calculate the number of moles of O2:
molar mass of O2 = 32.00 g/mol
moles of O2 = mass of O2 / molar mass of O2
moles of O2 = 1.0 g / 32.00 g/mol
moles of O2 ≈ 0.0313 mol
The balanced equation tells us that the stoichiometric ratio between S and O2 is 2:3. This means that for every 2 moles of S, 3 moles of O2 are needed to produce 2 moles of SO3.
3. Compare the moles of S and O2:
The ratio of moles of S to O2 is approximately 0.0312 mol : 0.0313 mol, which can be simplified to 1:1.
Since the ratio is 1:1, both reactants, S and O2, are in a 1:1 stoichiometric ratio. This means that the reactants will be completely consumed in the reaction, and there will be no excess reactant.
4. Calculate the maximum amount of SO3 that can be produced:
From the balanced equation, we know that 2 moles of SO3 are produced for every 2 moles of S.
moles of SO3 = moles of S
moles of SO3 = 0.0312 mol
To convert moles of SO3 to grams, use the molar mass of SO3:
molar mass of SO3 = 80.06 g/mol
mass of SO3 = moles of SO3 × molar mass of SO3
mass of SO3 = 0.0312 mol × 80.06 g/mol
mass of SO3 ≈ 2.489 g
Therefore, the maximum amount of SO3 that can be produced by the reaction 1.0 g of S with 1.0 g of O2 is approximately 2.49 g, which is closest to option B) 2.5 g.
To determine the maximum amount of SO3 that can be produced in this reaction, we need to identify the limiting reactant. The limiting reactant is the reactant that is completely consumed first and determines the maximum amount of product that can be formed.
To find the limiting reactant, we need to calculate the number of moles of each reactant using their respective molar masses.
Molar mass of S: 32.07 g/mol
Molar mass of O2: 32.00 g/mol
Using the given masses of S and O2 (1.0 g each), we can calculate the number of moles:
Number of moles of S = mass of S / molar mass of S = 1.0 g / 32.07 g/mol = 0.0312 mol
Number of moles of O2 = mass of O2 / molar mass of O2 = 1.0 g / 32.00 g/mol = 0.0313 mol
Now, we need to determine the stoichiometric ratio between S and O2 in the balanced equation. The balanced equation shows that 2 moles of S reacts with 3 moles of O2 to produce 2 moles of SO3.
Therefore, the ratio of moles of S to O2 is 2:3.
Since the ratio is not 1:1, we cannot directly compare the moles of S and O2. Instead, we need to find the limiting reactant by converting the moles of S and O2 to moles of SO3 that can be produced.
The maximum amount of SO3 that can be produced from S is:
0.0312 mol of S * (2 mol of SO3 / 2 mol of S) = 0.0312 mol of SO3
The maximum amount of SO3 that can be produced from O2 is:
0.0313 mol of O2 * (2 mol of SO3 / 3 mol of O2) = 0.0209 mol of SO3
Comparing the results, we see that O2 is the limiting reactant because the amount of SO3 produced from O2 (0.0209 mol) is smaller than the amount of SO3 produced from S (0.0312 mol).
Now, to calculate the maximum amount of SO3 produced in grams, we can use the molar mass of SO3.
Molar mass of SO3: 80.07 g/mol
Amount of SO3 produced from O2 = 0.0209 mol * 80.07 g/mol = 1.67 g
Therefore, the correct answer is D) 1.7 grams (rounded to the nearest tenth).
lets make it 100 g each..
moles S=100/32=3.33
moles O2=100/32=3.22
Rebalance the equation to
2/3 S+ O2 >> 2/3 SO3
so for each 3.33 moles of O, one needs 2/3(3.33) or 2.22 moles S, and gets 2.22 moles SO3 In this case, one has excess unreacted S.
2.22 moles SO3 will have a mass of
2.22*(32+3*16)=2.2*80=176 grams
Now scale that to 1/100, and you have it. Notice that I rounded a lot, you need to do it more accurately.