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
Moles of SO42- obtained = 8.3354 mol
Mass of SO42- obtained = 800.7g
Moles of H+ obtained = 8.3354mol
Mass of H+ obtained = 8.419g
Explanation
2FeS2(s) + 7O2(g) + 2H2O(l) --------> 2Fe2+(aq) + 4SO42-(aq) + 4H+(aq)
Stoichiometrically, 2moles of FeS2 give 4moles of SO42- and 4moles of H+
number of moles = mass / molar mass
given moles of FeS2 = 500g/ 119.97g/mol = 4.1677mol
moles of SO42- 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 SO42- obtained = 8.3354mol × 96.06g/mol = 800.7g
mass of H+ obtained = 8.3354mol × 1.01g/mol = 8.419g
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