Consider the reaction between reactants S & O2
2S(s) + 3 O2(g) ==> 2 SO3(g)
If a reactant vessel initially contains 5 mol S and 9 mol O2, how many moles of S, O2 and SO3 will be in the reaction vessel once the reactants have reached as much as possible? Assume 100% actual yield
Lost again
You do this as two simple stoichiometry problems, then take the smaller value.
5 mols S x (2 moles SO3/2 moles S) = 5 moles SO3 formed as one scenario.
9 moles O2 x (2 moles SO3/3 moles O2) = 6 moles SO3 formed as the second scenario.
Obviously both answers can't be correct; the smaller value is ALWAYS the correct one in limiting reagent problems (limiting reagent problems are those in which BOTH (or more if there are more than two) reactants are given and the reagent producing the smaller value is the limiting reagent. So 5 moles of S will react with the amount of O2 required and some O2 will remain un-reacted. Now that you know S is the limiting reagent, then
moles S after rxn will be zero.
moles SO3 will be 5.
moles O2 remaining is 9(initial)-amount used in the reaction. How much is used.
5moles S x (3 moles O2/2 moles S) = 7.5 used so 9-7.5 = 1.5 moles oxygen un-reacted to go with the zero moles S and 5 moles SO3.
no actually rodney is correct thank u i had to submit online and rodney was right not dr bob
The Answer submited by Dr.Bob222 is inncorrect. The limiting reagent is O2.There will be 1 mole of S remaining after the rxn. And thete will be 0 moles of 02 remaining(it is the limiting reagent so it will all be used in rxn). The final product will consist of 6 moles of SO3.
ya rodney u r dumb lol
wompus
Looked at both the responses, O2 is not the limiting reactant because it produces more reactants than S. The limiting reactant is the one that produces less product, so S is the limiting reactant cause it produces 1 less mole of SO3 than O2. DR BOB IS CORRECT!!!!
To determine the number of moles of S, O2, and SO3 in the reaction vessel once the reactants have reached as much as possible, you can use the concept of limiting reactants and stoichiometry.
1. Calculate the number of moles of S and O2 available:
- The initial number of moles of S is given as 5 mol.
- The initial number of moles of O2 is given as 9 mol.
2. Identify the limiting reactant:
To determine the limiting reactant, calculate the number of moles of SO3 that can be produced from each reactant, assuming complete reaction. This is done using the stoichiometric coefficients in the balanced equation.
- For S: According to the balanced equation, 2 moles of S react to produce 2 moles of SO3. Therefore, the number of moles of SO3 that can be produced from 5 moles of S is (5 mol S / 2 mol S) * (2 mol SO3 / 2 mol S) = 5 mol SO3.
- For O2: According to the balanced equation, 2 moles of S react with 3 moles of O2 to produce 2 moles of SO3. Therefore, the number of moles of SO3 that can be produced from 9 moles of O2 is (9 mol O2 / 3 mol O2) * (2 mol SO3 / 2 mol S) = 6 mol SO3.
Since the number of moles of SO3 that can be produced is higher for O2 (6 mol) compared to S (5 mol), O2 is the limiting reactant.
3. Calculate the number of moles of SO3 produced:
Since O2 is the limiting reactant, the number of moles of SO3 produced is equal to the number of moles of SO3 that can be produced from O2, which is 6 mol.
4. Calculate the remaining moles of S and O2:
To determine the remaining moles of S and O2 after the reaction, use the stoichiometry of the balanced equation.
- The remaining moles of S can be calculated as (5 mol S - 2 mol SO3 produced) = 3 mol S.
- The remaining moles of O2 can be calculated as (9 mol O2 - (3 mol SO3 produced * (3 mol O2 /2 mol SO3)) = 7.5 mol O2.
Therefore, once the reactants have reached as much as possible, there will be:
- 3 moles of S
- 7.5 moles of O2
- 6 moles of SO3 in the reaction vessel assuming 100% actual yield.