The mass of sulfur that has to burn to produce 4.5L SO2 at 300 c and 101 kPa is 3.05 grams (which is around answer B)
calculation
equation for reaction =S+ O2 = SO2
by use the ideal gas equation find the moles of SO2 formed
Ideal gas equation = PV=nRT where
P(pressure)= 101 KPa
V(volume)= 4.5 L
n(number of moles)=?
R(gas constant)= 8.314 L.Kpa/mol.k
T(temperature) = 300+ 273 =573 K
by making n the formula of the subject
n = PV/RT
n= (101 kpa x4.5 L) / ( 8.314 l.KPa/Mol.K x 573 K) = 0.0954 moles
by use of mole ratio between S to SO2 which is 1:1 the moles of SO2 is also 0.0954 moles
mass of SO2 = moles x molar mass
the molar mass of SO2 = 32 g/mol
therefore mass = 32 g/mol x 0.0954 = 3.05 grams(answer B)
