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?

2 answers

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.
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