What mass of sulfur has to burn to produce 4.5L SO2 at 300°C and 101 kPa in the following reaction?
S(s) + O2(g) -> SO2(g)
A. 3.07 g S
B. 41.0 g S
C. 13.5 g S
D. 68.8 g S
You can't add these up? Look on your periodic table and add the atomic masses. S + O2 ==> SO2
How many mols SO2 do you want? Use PV = nRT and solve for n.
Using the coefficients in the balanced equation, convert mols SO2 to mols S, then grams S = mols S x molar mass S.
Sorry, I answered one question twice. Ignore the first part.
To solve this problem, we need to use the ideal gas law and the stoichiometry of the reaction.
First, let's determine the number of moles of SO2 produced. We can use the ideal gas law equation:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Given:
Pressure (P) = 101 kPa
Volume (V) = 4.5 L
Temperature (T) = 300°C = 300 + 273 = 573 K
Using the ideal gas law equation, we can rearrange it to solve for the number of moles (n):
n = PV / RT
Plugging in the values:
n = (101 kPa) * (4.5 L) / (0.0821 L·atm/mol·K) * (573 K)
n ≈ 0.89 mol
Next, we need to consider the stoichiometry of the reaction. From the balanced equation:
1 mol of S reacts with 1 mol of O2 to produce 1 mol of SO2.
Therefore, the number of moles of S required will be the same as the number of moles of SO2 produced, which is approximately 0.89 mol.
Now, let's convert the moles of S to grams. We can use the molar mass of S:
Molar mass of S = 32.06 g/mol
Grams of S = moles of S * molar mass of S
Grams of S = 0.89 mol * 32.06 g/mol
Grams of S ≈ 28.5 g
Therefore, the mass of sulfur that needs to burn to produce 4.5 L of SO2 at 300°C and 101 kPa is approximately 28.5 grams.
None of the given answer choices match this result exactly. However, the closest option is C. 13.5 g S.