16. What mass of sulfur has to burn to produce 4.5L SO2 at 300°C and 101 kPa in the following reaction? A. 41.0 g S B. 3.07 g S C. 68.8 g S D. 13.5 g S

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)