In one process 4 grams of sulphur was burnt in 48 decimetre cube of oxygen measured at room temperature. What is the limiting reactant in this reaction?

Answer 1

Here is the solution but see the note that follows.
S + O2 ==> SO2
mols S = grams/atomic mass = 4/32 = 0.125
volume O2 = 48 dm^3 = 48 L and
mols O2 = 48/22.4 = 2.14
Look at the equation. That tells you that 1 mol S reacts exactly with 1 mol O2. So you have 0.125 mol S and ALL of that will react with the 0.125 mols O2 so S is the limiting reagent and some O2 will remain un-reacted.

NOTE: It is now known that S exists as S8. Technically, the correct equation is S8 + 8O2 ==> 8SO2. The correct answer still is S as the limiting reagent but the first equation is easier to solve.

Answer 2

In one process 4.0g of Sulphur was burnt in 48.0dm3 of oxygen measured at r.t.p. calculate the volume of Sulphur dioxide formed at r.t.p