Question

Sulfuric acid (H2SO4) is prepared commercially from elemental sulfur using the contact process. In a typical sequence of reactions, the sulfur is first burned: S + O2 → SO2 , then it is converted to SO3 using a catalyst: 2 SO2 + O2 → 2 SO3 . The resulting SO3 is reacted with water to produce the desired product: SO3 + H2O → H2SO4 . How much sulfuric acid could be prepared from 83 moles of sulfur? Answer in units of g.

Answer 1

I believe the end result is still 83 moles since there is never an amount of sulfur atoms added to the initial amount, but rather oxygen and water is repeatedly added to it. To find it's weight, first find the molar mass of H2SO4:

H2 + S + O4 = 2.00 + 32.1 + 64.0 = 98.1 g/mol

and mass = (98.1 g/mol)(83 mol) = 8142.3 g

rounded to 8.1 x 10^3 g assuming 100% yield?

Answer 2

Answer:

$8134gH_{2}SO_{4}$

Explanation:

From the first reaction:

$S+O_{2}=SO_{2}$

Finding the number of moles using stoichiometry:

$83molesS*\frac{1molSO_{2}}{1molS}=83molesSO_{2}$

Using stoichiometry in the second reaction:

$_{2}SO_{2}+O_{2}=_{2}SO_{3}$

$83molesSO_{2}*\frac{2molesSO_{3}}{2molesSO_{2}}=83molesSO_{3}$

From the third reaction:

$SO_{3}+H_{2}O=H_{2}SO_{4}$

$83molesSO_{3}*\frac{1molH_{2}SO_{4}}{1molSO_{3}}=83molesH_{2}SO_{4}$

Finding the mass in grams of $H_{2}SO_{4}$:

$83molesH_{2}SO_{4}*\frac{98gH_{2}SO_{4}}{1molH_{2}SO_{4}}=8134gH_{2}SO_{4}$