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Iodine can be prepared by the following reactions. `2NaIO_(3) + 5NaSO_(3) rarr 2NaSO_(4) + 2Na_(2) SO_(4) + H_(2)O + I_(2)` How much `NaHSO_(3)` is required to produce `381 g` of `I_(2)`?A. `156.0 g`B. `390.0 g`C. `520.0 g`D. `780.0 g |
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Answer» Correct Answer - D Mol of `I_(2) = (381)/(254) = 1.5 "mol" I_(2)` `(1.5 "mol" I_(2)) ((5 "mol" NaHSO_(3))/("mol" I_(2))) ((104 g NaHSO_(3))/("mol" NaHSO_(3)))` `implies 1.5 xx 5 xx 104 = 780 g "of" NaHSO_(3)` |
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