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?
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
Well, it seems like the sulfur and oxygen got into quite a heated argument! But let's see who the real limiting reactant is.
To determine the limiting reactant, we first need to find the moles of sulfur and oxygen in the given quantities.
The molar mass of sulfur (S) is 32 g/mol, so 4 grams of sulfur is equal to (4 g / 32 g/mol) = 0.125 moles of sulfur.
The molar mass of oxygen (O₂) is 32 g/mol as well. We need to convert the volume of oxygen to moles using the ideal gas law, but since the volume is measured at room temperature, we can assume the pressure is constant.
On the other hand, we have the volume of oxygen (48 dm³). Since 1 dm³ is equal to 1000 cm³, we can convert it to cm³: 48 dm³ * 1000 cm³/dm³ = 48000 cm³.
To convert cm³ to liters, we divide it by 1000: 48000 cm³ / 1000 = 48 L.
Using the ideal gas law, we can find the moles of oxygen: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
Since we're assuming constant pressure at room temperature, n = PV/RT. The gas constant (R) is approximately 0.0821 L·atm/(mol·K). At room temperature, we'll assume 298 K.
n = (48 L * 1 atm) / (0.0821 L·atm/(mol·K) * 298 K) ≈ 1.964 moles of oxygen.
Now that we have the moles for both sulfur and oxygen, we compare their ratios in the balanced chemical equation of the reaction. The equation for the combustion of sulfur is:
S + O₂ -> SO₂
From the equation, we can see that the ratio of S to O₂ is 1:1. This means that to react completely, we would need 0.125 moles of sulfur and 0.125 moles of oxygen.
However, we have 1.964 moles of oxygen, which is way more than the required amount. So, in this case, oxygen is the excess reactant, and sulfur is the limiting reactant.
Looks like sulfur needed to put its foot down and say, "Enough is enough!"
To determine the limiting reactant in this reaction, we need to compare the amount of reactants used with their stoichiometric ratio.
1. Start with the balanced chemical equation for the combustion of sulfur:
S + O₂ -> SO₂
2. Convert the mass of sulfur to moles using the molar mass of sulfur:
Molar mass of S = 32.06 g/mol
Moles of S = 4 g / 32.06 g/mol = 0.125 mol
3. Convert the volume of oxygen to moles using the ideal gas law:
Volume of O₂ = 48 dm³ = 48 L
Molar volume of a gas at room temperature and pressure (STP) is 22.4 L/mol.
Moles of O₂ = 48 L / 22.4 L/mol = 2.143 mol
4. Determine the stoichiometric ratio of sulfur to oxygen:
From the balanced equation, we can see that 1 mol of S reacts with 1 mol of O₂ to produce 1 mol of SO₂.
5. Compare the moles of S and O₂ in the stoichiometric ratio:
The moles of sulfur (0.125 mol) are less than the moles of oxygen (2.143 mol).
Therefore, sulfur is the limiting reactant in this reaction as it's present in lesser quantity compared to oxygen.
To determine the limiting reactant in the reaction, we need to compare the amount of each reactant with the stoichiometry of the balanced chemical equation.
First, let's consider the balanced equation for the combustion of sulfur:
S + O2 -> SO2
From this equation, we can see that the stoichiometric ratio between sulfur and oxygen is 1:1. That means for every 1 mole of sulfur, we need 1 mole of oxygen.
Now, let's calculate the number of moles of sulfur and oxygen in the given quantities:
Mass of sulfur = 4 grams
Molar mass of sulfur (S) = 32 grams/mol
Number of moles of sulfur = (4 grams) / (32 grams/mol) = 0.125 mol
Volume of oxygen = 48 decimetre cube = 48 dm^3
At room temperature and pressure, 1 mole of any gas occupies 22.4 dm^3
Number of moles of oxygen = (48 dm^3) / (22.4 dm^3/mol) = 2.14 mol
From the above calculations, we can see that there is an excess of oxygen compared to sulfur. We need only 0.125 moles of oxygen for the complete reaction, but we have 2.14 moles of oxygen available.
Therefore, the limiting reactant in this reaction is sulfur (S), as there is not enough sulfur to react completely with the excess oxygen.