Exactly 75.00 mL of a 333.3 ppm (w/v) sodium sulfite (MM=126.041 g/mol), solution is treated with 150.0 mL of 3.894×10^-3 M perchloric acid

Question

Exactly 75.00 mL of a 333.3 ppm (w/v) sodium sulfite (MM=126.041 g/mol), solution is treated with 150.0 mL of 3.894×10^-3 M perchloric acid (MM= 100.457 g/mol), which results in the formation of sulfur dioxide (MM=64.064 g/mol), water (MM=18.015 g/mol), and sodium perchlorate (MM=122.438 g/mol). What is the ppm (w/v) of the excess reactant once the reaction is complete?

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Mít Mít 3 years 2021-07-14T06:42:34+00:00 1 Answers 28 views 0

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  1. quangkhai
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    2021-07-14T06:44:04+00:00
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    Answer:

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