Sulfuric Acid Section – Daily Question for November 24th

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This post was last edited by “I Love Chemical Engineering” on November 24, 2011, at 15:13. 1. Please hide your replies; editing after posting is not possible! Delete directly without hiding! 2. Method for hiding: http://bbs.hcbbs.com/thread-492556-1-1.html. 3 Points will be added for correct answers; everyone who participates will receive a reward. A certain sulfuric acid factory acquired a batch of pyrite, and the process used to produce sulfuric acid is the contact process. The factory’s laboratory took 10.0 grams of ore sample, fully calcined it in an oxygen stream, and then weighed it; the mass of the residue was 7.60 grams. Calculate the mass fraction of FeS2 in the ore (the impurity is silicon dioxide). A 72% B 78% C 82% D 86%. In the previous question, sulfur is conserved: it was present in 98% concentrated sulfuric acid initially and remains at 98% at the end. Moreover, for H2SO4*SO3, the ratio between H2SO4 and SO3 is 1:1, which means the amount of sulfuric acid produced doubles. Since there is also 2% water present, this amount needs to be taken into account as well. 2% water corresponds to 20 grams, and 20/18 moles of water; accordingly, 20/18 moles of sulfuric acid are produced. Thus, we get a total of 980/98 + 20/18 moles of pure sulfuric acid! Considering the balance of sulfur atoms in H2SO4*SO3, the total amount of sulfuric acid is 2*(980/98 + 20/18) = 22.222 moles. Multiplying this by the mass and dividing by the concentration gives 22.22*98/0.98 = 2200 grams, or 2.222 kilograms