Answer / kang chuen tat (malaysia - pen

Answer 43 - Basis = 100 mol / h. For inlet, mole % of carbon dioxide = 8 % and mole % of nitrogen = 100 - 8 = 92 % then mole ratio = 8 / 92 = 0.086957. For outlet, mole % of nitrogen = 92 %, mole % of carbon dioxide = 8 % (1 - 0.65) = 2.8 %, then mole ratio = 2.8 / 92 = 0.030435. (b) When x = 0.0000427, y = 1640 x = 1640 (0.0000427) = 0.070028, then mole ratio = y / (1 - y) = 0.0753. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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