posted by Monique .
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
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.
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.
Dr.Bob222 is right. Rodney, your answer is wrong. I just did this problem online and the correct answers were the one that Dr.Bob22 gave.
Agreed. Dr. Bob is correct.
ya rodney u r dumb lol
i don't understand can you explain about that