Question

3. The following reaction describes the oxidation of pyrite (FeS2) by oxygen in the presence of water:

2FeS2 (s) + 7O2 (g) + 2H2O → 2Fe2+ + 4SO42- + 4H+

Calculate the amount of sulfate and H+ expressed both in number of moles and grams when 500 grams of pyrite react according to the above reaction and there is enough oxygen for the reaction to go to completion.

Answer 1

Answer

Moles of SO4^2- obtained = 8.3354 mol

Mass of SO4^2- obtained = 800.7g

Moles of H⁺ obtained = 8.3354mol

Mass of H⁺ obtained = 8.419g

Explanation

2FeS2(s) + 7O2(g) + 2H2O(l) --------> 2Fe²⁺(aq) + 4SO4^2-(aq) + 4H⁺(aq)

Stoichiometrically, 2moles of FeS2 give 4moles of SO4^2- and 4moles of H⁺

number of moles = mass / molar mass

given moles of FeS2 = 500g/ 119.97g/mol = 4.1677mol

moles of SO4^2- obtained = (4/2)× 4.1677mol = 8.3354mol

moles of H⁺ obtained = (4/2)×4.1677mol = 8.3354mol

mass = number of moles × molar mass

mass of SO4^2- obtained = 8.3354mol × 96.06g/mol = 800.7g

mass of H⁺ obtained = 8.3354mol × 1.01g/mol = 8.419g