In one process, 4.0 grams of sulphur was brunt in 48.0 decimeter cube of oxygen measured at RTP.
(A) What is the limiting reactant in this reaction?
(B) Calculate the volume of sulphur dioxide formed at RTP.
Another limiting reagent problem Follow the directions in your posts above. Post your work if you get stuck and I can help you through it.
What is RTP? STP maybe?
Please show what you have done.
RTP is Room Temperature & Pressure. Room T is considered 298 K.
To determine the limiting reactant in a chemical reaction, you need to compare the stoichiometric ratios of the reactants and see which one is present in the smallest amount relative to the balanced equation.
(A) Let's start by writing and balancing the chemical equation for the reaction between sulfur (S) and oxygen (O2) to form sulfur dioxide (SO2):
2S + O2 -> 2SO2
In this equation, the stoichiometric ratio between S and O2 is 2:1. This means that for every 2 moles of sulfur, we need 1 mole of oxygen to react completely.
Now, let's convert the given mass of sulfur (4.0 grams) and volume of oxygen (48.0 dm^3) into moles. We'll use the molar mass of sulfur and the ideal gas law to do this.
The molar mass of sulfur is 32.06 g/mol, so the moles of sulfur can be calculated as:
moles of sulfur = mass of sulfur (g) / molar mass of sulfur (g/mol)
moles of sulfur = 4.0 g / 32.06 g/mol
moles of sulfur ≈ 0.125 mol
To convert the volume of oxygen to moles, we'll use the ideal gas law equation:
PV = nRT
Assuming RTP (room temperature and pressure) is 298 K and 1 atm, the ideal gas law equation simplifies to:
V = nRT / P
V is the volume of the gas (48.0 dm^3), n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), T is the temperature in Kelvin, and P is the pressure (1 atm).
Substituting the given values into the ideal gas law equation:
n = V × P / (R × T)
n = 48.0 dm^3 × 1 atm / (0.0821 L·atm/(mol·K) × 298 K)
n ≈ 1.958 mol
Now that we have the number of moles for both reactants, we can compare the stoichiometric ratios. The ratio of sulfur to oxygen is 2:1. We have 0.125 moles of sulfur and 1.958 moles of oxygen.
To determine the limiting reactant, we compare the moles of each reactant to their stoichiometric ratio. Since the stoichiometric ratio is 2:1, we can multiply the moles of sulfur by 2 to compare them correctly.
0.125 mol × 2 = 0.250 mol
Comparing 0.250 mol of sulfur to 1.958 mol of oxygen, we see that sulfur is present in a smaller amount relative to their stoichiometric ratio. Therefore, sulfur is the limiting reactant in this reaction.
(B) Now that we know sulfur is the limiting reactant, we can use the stoichiometry to calculate the volume of sulfur dioxide formed. According to the balanced equation, 2 moles of sulfur react to form 2 moles of sulfur dioxide (SO2).
Since we have determined that 0.125 mol of sulfur is present, we expect to obtain the same number of moles of sulfur dioxide.
To convert moles of sulfur dioxide to volume, we can use the ideal gas law:
V = nRT / P
Using the same conditions as before (RTP), and assuming the volume of sulfur dioxide is V, the equation becomes:
V = 0.125 mol × 0.0821 L·atm/(mol·K) × 298 K / (1 atm)
V ≈ 3.69 dm^3
Therefore, the volume of sulfur dioxide formed at RTP is approximately 3.69 dm^3.