Exam 3 Fall 09 question 11

Exam 3 Fall 09 question 11 - n(rnol SO2 = 1.437 x lo4 mol I...

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n (mol) I2 = 0.0200 L x 0'01017 mol - 2.034 x lo4 mol 1, initially 1 L solution 0.0105 mol S,O;- 1 mol I2 n (mol) I, = 0.01 137 L x x 1 L solution 2 mol s,o,~- = 5.97 x rnol I2 (unreacted with SO2) n (rnol) I2 = 2.034 x lod rnol - 5.97 x 10" rnol = 1.437 x
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Unformatted text preview: n (rnol) SO2 = 1.437 x lo4 mol I, x mO1 = 2.874 x lo4 mol SO, 1 rnol I2 [700. ton x (1 a m 6 0 ton)] (0.500 L) n (mol) air = = 1.80 x 1 0-, mol air (0.0821 L , atm/mol . K) (273 + 2 5 ) ~ 2.874 x lo4 rnol SO2 volume%S02 = mol%S02 = x 100 = 1.59% 1.80 x lo-' rnol air...
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