The burning of sulfur can be described by the following equation: S(s) + O2(g) → SO2(g) What volume of sulfur dioxide gas will be released at 25°C and 101.3 kPa when 8.00 g of sulfur is burnt? - HSC - SSCE Chemistry - Question 8 - 2001 - Paper 1

From the balanced equation, we see that 1 mole of S produces 1 mole of SO2. Therefore, the moles of SO2 produced will also be approximately 0.249 mol.

To calculate the volume of SO2 produced at 25°C (which is 298 K) and a pressure of 101.3 kPa, we can use the ideal gas law:

$PV = nRT$

Where:

• P is the pressure (101.3 kPa)
• V is the volume (what we want to find)
• n is the number of moles of gas (0.249 mol)
• R is the ideal gas constant (8.314 J/(mol K), or 8.314 kPa L/(mol K))
• T is the temperature in Kelvin (298 K)

Rearranging gives us:

$V = \frac{nRT}{P}$ Now substituting the values:

$V = \frac{0.249 \text{ mol} \times 8.314 \text{ kPa L/(mol K)} \times 298 \text{ K}}{101.3 \text{ kPa}} \approx 6.12 \text{ L}$