Question

2S(s)+3O₂(g)→2SO₃(g)

A. If a reaction vessel initially contains 7 molS and 9 mol O 2, how many moles of S will be in the reaction vessel once the reactants have reacted as much as possible? (Assume 100% actual yield.)

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

Show all work and explain each step please!

Answer 1

Given reaction is:

                                                          2S(s)    +      3O₂(g)   →      2SO₃(g)

a) So from the above balanced chemical equation we can understand that:

2 moles of sulphur is reacting with                    -------------- 3 moles of oxygen.

If we take initially 7 moles of sulphur, then it requires -----    ? moles of oxygen

= (3 moles of oxygen * 7 moles of sulphur) / 2 moles of sulphur

= 10.5 moles of oxygen is required but we have taken only 9 moles which is less than required, hence oxygen is the limiting reagent.

2 moles of sulphur is reacting with                    -------------- 3 moles of oxygen.

? moles of sulphur                          ----------------------------- reacts with 9 moles of oxygen

= (9 moles of oxygen * 2moles of sulphur)/ 3 moles of oxygen

= 6 moles of sulphur

The excess amonut of sulphur remained after the reaction is : 7moles-6moles

                                                                                                = 1mole of sulphur remains after the reaction.

b)From the above balanced chemical equation we can understand that:

2 moles of sulphur is reacting with                    -------------- 3 moles of oxygen.

If we take initially 7 moles of sulphur, then it requires -----    ? moles of oxygen

= (3 moles of oxygen * 7 moles of sulphur) / 2 moles of sulphur

= 10.5 moles of oxygen is required but we have taken only 9 moles which is less than required, hence oxygen is the limiting reagent.

Hence oxygen wont remain after the reaction. It will be consumed completely during the reaction.