Question

Consider the reaction between reactants S and O2: 2S(s)+3O2(g)→2SO3(g)

1. If a reaction vessel initially contains 5 molS and 9 molO2, how many moles of S will be in the reaction vessel once the reactants have reacted as much as possible? (Assume 100% actual yield.)

Express your answer using two significant figures.

2. How many moles of O2 will be in the reaction vessel once the reactants have reacted as much as possible? (Assume 100% actual yield.)

Express your answer using two significant figures.

3. How many moles of SO3 will be in the reaction vessel once the reactants have reacted as much as possible? (Assume 100% actual yield.)

Express your answer using two significant figures.

Answer 1

Ans. Balanced reaction:          2S(s) + 3 O2(g) -------> 2 SO3(3)

Stoichiometry: 2 mol S reacts with 3 mol O2 to produce 2 mol SO3.

So, theoretical stoichiometric ratio of reactant = S : O₂ = 2 : 3

#1. Given,

            Initial moles of S = 5 mol

            Initial moles of O2 = 9 mol

Experimental ratio of reactants = S: O₂ = 5 : 9

Now,

Convert the experimental ratio by multiplying (or dividing) by a factor such that at least both the ratios have one common digit.

Dividing experimental ratio by factor 3, we get -

S: O₂ = (5 : 9) / 3 = 1.66 : 3              - corrected experimental ratio

Note that multiplying or dividing a ratio by a factor does not affect the ration.

Compare theoretical stoichiometric ratio and corrected experimental ratio of reactants, the moles of S (1.66 mol) in experimental ratio is less than that of theoretical ratio (2 mol) while keeping moles of O₂ constant. So, S-is the limiting reactant.

Being the limiting reactant, all of S will be converted into product. Therefore-

            Amount of S after completion (100%) of reaction = 0.0 (zero).

#2. See stoichiometry, 2 moles consumed 3 mol O2.

So,

Moles of O₂ consumed during reaction = (3/ 2) x moles of S consumed

                                                = (3/ 2) x 5 mol

                                                = 7.5 mol

Remaining moles of O2 = Initial moles – moles consumed in reaction

                                    = 9 mol- 7.5 mol

                                    = 1.5 mol

#3. See stoichiometry, 2 moles S produces 2 mol SO3.

So,

Moles of SO3 in vessel = moles of S consumed

                                    = 5.0 mol