Elementary Principles 142 - 5.48 (cont'd) c. SO 3 leaving...

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5- 30 5.48 (cont’d) c. SO leaving converter: (0.6970 + 0.4687) kmol =1.156 kmol 3 1.156 kmol SO min kmol H SO kmol SO kg H SO kmol kg H SO 32 4 3 24 ⇒= 1 1 98 1133 . Sulfur in ore: 0.683 kmol FeS kmol S kmol FeS kg S kmol kg S 2 2 23 2 1 438 . . = 259 . . kg H SO 43.8 kg S kg H SO kg S = 100% conv.of S: 0.683 kmol FeS kmol S kmol FeS kmol H SO 1 kmol S kg kmol kg H SO 43.8 kg S kg H SO kg S 2 2 21 9 8 1339 306 = . . . The sulfur is not completely converted to H 2 SO 4 because of (i) incomplete oxidation of FeS 2 in the roasting furnace, (ii) incomplete conversion of SO 2 to SO 3 in the converter. 5.49 NO 2 2 a. n P1 . 0 0 V RT 2.00 atm 2.00 L 473K 0.08206 L atm mol - K 0.103 mol NO 0 gauge 0 2 = + = = di bg b g b g b. nm o l N O 12 = , n mol N O 22 4 = py P n nn P NO NO 1 == + F H G I K J , p n PK n nn n P 2 p 1 2 21 2 = + F H G I K J + Ideal gas equation of state ⇒=+ + = PV n n RT n n PV / RT 1 b g Stoichiometric equation each mole of N O present at equilibrium represents a loss of two moles of NO 2 from that initially present ⇒+ = n2 n 0 . 1 0 Solve (1) and (2)
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