in one process, 4.0g of sulfur was burnt in 48.0dm^3 of oxygen measured at r.t.p.
a) what is the limiting reactant in this reaction
b) calculate the volume of Sulfur dioxide formed at r.t.p
To determine the limiting reactant, we need to compare the moles of each reactant and see which one limits the reaction.
First, let's calculate the moles of sulfur and oxygen:
Moles of Sulfur = Mass of Sulfur / Molar mass of Sulfur
Moles of Sulfur = 4.0 g / 32.06 g/mol (molar mass of Sulfur)
Moles of Sulfur = 0.1248 mol
Moles of Oxygen = Volume of Oxygen / Molar volume of Oxygen at r.t.p.
Moles of Oxygen = 48.0 dm^3 / 22.4 dm^3/mol (molar volume of gas at r.t.p.)
Moles of Oxygen = 2.143 mol
The balanced equation for the reaction is:
Sulfur + Oxygen -> Sulfur Dioxide
From the equation, we can see that the mole ratio between Sulfur and Oxygen is 1:1. This means that 0.1248 moles of Sulfur will react with 0.1248 moles of Oxygen to form Sulfur Dioxide.
Since we have 2.143 moles of Oxygen and only 0.1248 moles of Sulfur, it is clear that Oxygen is the excess reactant and Sulfur is the limiting reactant.
To calculate the volume of Sulfur Dioxide formed at r.t.p., we can use the same mole ratio:
0.1248 moles of Sulfur Dioxide = 0.1248 moles of Sulfur
Using the ideal gas law, we can calculate the volume of Sulfur Dioxide:
Volume of Sulfur Dioxide = Moles of Sulfur Dioxide x Molar volume at r.t.p.
Volume of Sulfur Dioxide = 0.1248 mol x 22.4 dm^3/mol
Volume of Sulfur Dioxide = 2.79 dm^3
Therefore, the volume of Sulfur Dioxide formed at r.t.p. is 2.79 dm^3.