Answer to Question #68868 in General Chemistry for Laura

Question

You are overseeing a chemical plant production of 100 kilograms (kg) of sulfuric acid on a trial basis using the contact process. Determine the exact quantities of each substance involved assuming 100% yield for all reactions.

List in order each calculation and answer. The equations for the process are as follows:

Step 1. S(g) + O2(g) = SO2(g)
Step 2. SO2(g) + ½O2(g) = SO3(g) [This reaction requires a divanadium pentoxide catalyst.]
Step 3. SO3(g) + H2SO4(l) = H2S2O7(l)
Step 4. H2S2O7(l) + H2O(l) = 2 H2SO4(aq)

Atomic Masses: S = 32.0, O = 16.0, H = 1.00

You may estimate the quantity of the catalyst.

NOTE: No dilution is intended as these are masses, not solution concentrations.

Answer 1

We need to get material balance equation. For this we need firstly write sum of all reactions:
Full equation: S(g) + O2(g) + SO2(g) + ½O2(g) + SO3(g) + H2SO4(l) + H2S2O7(l) + H2O(l) = SO2(g) + SO3(g) + H2S2O7(l) + 2 H2SO4(aq)
Abbreviated equation: S(g) + 3/2O2(g) + H2SO4(l) + H2O(l) = 2 H2SO4(aq)
Now we calculate the number of mole-equivalents sulfuric acid:
nE(H2SO4)=m(H2SO4)2∙M(H2SO4)=100∙103g2∙98gmol=0.5102(kmol-equivalent)
Quantities of each substance:
m(S)=nE(H2SO4)∙1∙M(S)=0.5102∙32=16.326 (kg)
m(O2)=nE(H2SO4)∙1.5∙ M(O2)=0.5102∙1.5∙32=24.49 (kg)
m(H2SO4)=nE(H2SO4)∙1∙ M(H2SO4)=0.5102∙98=50 (kg)
m(H2O)=nE(H2SO4)∙1∙ M(H2O)=0.5102∙18=9.184 (kg)