Elementary Principles 338

# Elementary Principles 338 - 8.83 Basis: 100 mol solution 20...

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8-59 8.83 Basis: 100 mol solution 20 mol NaOH, 80 mol H 2 O r == 80 400 mol H O 20 mol NaOH mol H O mol NaOH 2 2 . Refs: NaOH(s), H O @25 C 2 l bg ° substance NaOH in mol HO i n k Jm o l NaOH in mol NaOH in in out out 2 nH sn lH rn ±± .. ± . 20 0 0 0 80 0 0 0 4 00 20 0 34 43 −− =− −← ± . Hr NaOH, 4.00 kJ mol 34 43 NaOH (Table B.11) Δ Hn H n H ii =−= = −× ∑∑ out in kJ 9.486 10 Btu kJ Btu () ( .) . . 20 34 43 688 6 10 6532 4 3 Q = + 10 20 0 40 00 80 0 18 01 132 3 3 . . Btu g g 2.20462 lb Btu lb product solution m m 8.84 Basis: 1 liter solution n HSO 2 4 2 4 24 1 L 8 g - eq 1 mol L 2 g - eq 4 mol H SO kg 1 mol kg H SO × F H G I K J = 0 09808 0 392 . . m total L 1.230 kg L kg solution 1 1230 . n 22 2 2 2 kg H O 1000 mol H O 18.02 kg H O mol H O = = 0 392 465 . ⇒= = = r n n 2 2 2 46.49 mol H O 4 mol H SO 11.6 mol H O mol H SO HS Oa q , =
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## This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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