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 need to go back and look at your posts. I worked this earlier IN GREAT DETAIL.

To determine the amount of S, O2, and SO3 in the reaction vessel once the reaction has reached completion, we can use the concept of stoichiometry.

First, let's analyze the balanced equation for the reaction:

2S(s) + 3O2(g) -> 2SO3(g)

According to the balanced equation, it takes 2 moles of S react with 3 moles of O2 to produce 2 moles of SO3. This means that the ratio of S to O2 to SO3 is 2:3:2.

Given that the initial amounts of S and O2 are 5 mol and 9 mol, respectively, we can calculate the limiting reactant.

1. Calculate the moles of S: 5 mol
2. Calculate the moles of O2: 9 mol

Next, compare the moles of S and O2 in the ratio of 2:3 to determine the limiting reactant.

3. Ratio of S to O2: 2 mol S / 3 mol O2
4. Moles of O2 required to react with available S: (5 mol S) * (3 mol O2 / 2 mol S) = 7.5 mol O2

Since we only have 9 mol of O2 available, it is in excess, and S is the limiting reactant.

Now, we can use the limiting reactant to determine the moles of SO3 produced.

5. Determine the moles of SO3 produced: (5 mol S) * (2 mol SO3 / 2 mol S) = 5 mol SO3

Therefore, once the reactants have reached as much as possible, there will be 0 moles of S and 4 mol of O2 remaining, while 5 mol of SO3 will be produced in the reaction vessel, assuming 100% actual yield.