Question

A mixture of carbon and sulfur has a mass of 9.0 g. Complete combustion with excess O2 gives 26.0 g of a mixture of CO2 and SO2.

Find the mass of sulfur in the original mixture.

Answer 1

m = 9 g mix

m = 26 g of CO2 and SO2

find mix of each

C + O2 = CO2

S + O2 = SO2

then

MW of C = 12 and MW CO2 = 44

MW of S = 32.065 MW of SO2 = 64

then...

state mol balances:

mass C = molC/MWC

mass of S = molS/MWS

then

mass C + mass S = total mass

molC/MWC + molS/MWS = 9

molC/12 + molS/32 = 9 (EQUATION 1)

then

for oxide mix

mass CO2 = molCO2/MWCO2

mass of SO2 = molSO2/MWSO2

equation:

mass of CO2 + mass of SO2 = total mass

molCO2*MWCO2 + molSO2*MWSO2 = 26

molCO2*44+ molSO2*64= 26 (EQUATION 2)

from eqn 1 and 2

molC*12 + molS*32 = 9 (EQUATION 1)

molCO2*44+ molSO2*64= 26 (EQUATION 2)

important note: 1 mol of C will form 1 mol of CO2 and 1 mol of S forms 1 mol of SO2 then

assume x as mol of C, CO2 and y as mol of S, SO2

x*12 + y*32 = 9

x*44+ y*64= 26

solve for x

x = (9-32y)/12

substitutein (29

(9-32y)/12*44+ y*64= 26

33-117.3y + 64y = 26

(64-117.3)y = 26-33

y = (26-33)/((64-117.3)) = 0.13133 mol of S and SO2

for x

x = (9-32y)/12 = x = (9-32*0.13133 )/12 = 0.399786 mol of C and CO2

mass of sulfur original

mass of S = mol*MW = 0.13133 *32 = 4.20256 g of S