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

According to the reaction, 1 mole of sulfur produces 1 mole of sulfur dioxide (SO2). Therefore, 0.249 moles of sulfur will produce 0.249 moles of SO2 as well.

To find the volume of gas using the ideal gas law, we use:
$PV = nRT$
where:

• $P$ = pressure in kPa (101.3 kPa),
• $V$ = volume in liters,
• $n$ = number of moles (0.249 moles),
• $R$ = ideal gas constant (8.314 L·kPa/(K·mol)),
• $T$ = temperature in Kelvin (25°C = 298 K).

Rearranging for volume, we have:
$V = \frac{nRT}{P}$
Substituting the values:
$V = \frac{0.249 ext{ moles} \times 8.314 ext{ L·kPa/(K·mol)} \times 298 ext{ K}}{101.3 ext{ kPa}} \approx 6.12 ext{ L}$