Question

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

Answer 1

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)