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
To find the mass of sulfur needed to produce 4.5 liters of SO2, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (101 kPa)
V = volume (4.5 L)
n = moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (300°C + 273.15 K = 573.15 K)
First, we need to calculate the number of moles of SO2 using the given volume and temperature.
n(SO2) = PV / RT
= (101 kPa) * (4.5 L) / (0.0821 L·atm/(mol·K) * 573.15 K)
≈ 0.3643 moles SO2
Now, we can see from the balanced equation that each mole of SO2 is produced from one mole of S. Therefore, the number of moles of S needed is also 0.3643 moles.
Finally, we can calculate the mass of sulfur using the molar mass of S (32.07 g/mol)
Mass of S = moles of S * molar mass of S
= 0.3643 moles * 32.07 g/mol
≈ 11.68 g
Therefore, the correct answer is option C. 13.5 g S (rounded to two significant figures).