Asked by Chasm
Sulfuric acid is prepared commercially from elemental sulfur using the contact process. In a typical sequence of reactions, the sulfur is first burned: S + O2 yields SO2, then it is converted to SO3 using a catalyst: 2SO3 + O2 yields 2SO3. The resulting SO3 is reacted with water to produce the desired product: SO3 + H2O yields H2SO4. How much sulfuric acid could be prepared from 49 moles of sulfur?
Answer in units of g
Answer in units of g
Answers
Answered by
DrBob222
S + O2 = SO2
2SO2 + O2 ==> 2SO3
SO3 + H2O ==> H2SO4
This is a simple stoichiometry problem made a little more complicated because you must deal with three reactions instead of one. You COULD work three problems (For example: 49 mols S will prepare x mols SO2, x mols SO2 will prepare y mol SO3, y mols SO3 will prepare z mols H2SO4. A little shorter way is to do all of the equations at once this way.
49 mol S x (1 mol SO2/1 mol S) x (2 mol SO3/2 mols SO2) x (1 mol H2SO4/1 mol SO3) = 49 mol H2SO4 when all of the numerators and denominators are canceled.
Then g H2SO4 = mols H2SO4 x molar mass H2SO4. The answer is about 49mols x 98g/mol = about 4802 g H2SO4.
2SO2 + O2 ==> 2SO3
SO3 + H2O ==> H2SO4
This is a simple stoichiometry problem made a little more complicated because you must deal with three reactions instead of one. You COULD work three problems (For example: 49 mols S will prepare x mols SO2, x mols SO2 will prepare y mol SO3, y mols SO3 will prepare z mols H2SO4. A little shorter way is to do all of the equations at once this way.
49 mol S x (1 mol SO2/1 mol S) x (2 mol SO3/2 mols SO2) x (1 mol H2SO4/1 mol SO3) = 49 mol H2SO4 when all of the numerators and denominators are canceled.
Then g H2SO4 = mols H2SO4 x molar mass H2SO4. The answer is about 49mols x 98g/mol = about 4802 g H2SO4.
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