1.

100 grams of each P_(4)O_(6) and KMnO_(4)^(-) were mixed in hydrochloric acid solution to from H_(3)PO_(4) and MnCl_(2). Which reagentis left unreacted and how much of it is left ?

Answer»

Solution :`MN^(2+) + 5E^(-) rarr Mn^(2+)`
Thus milliequivalents of `KMnO_(4) = (100 xx 5 x 1000)/(158) = 3164.56`
`(P^(3+))_(4) rarr 4P^(5+) + 8e^(-)`
Thus MILLI equivalents of `P_(4)O_(6) = (100 xx 8 xx 1000)/(220) = 3638`
Therefore `P_(4)O_(6)` is excess. Excess milliequivalents of `P_(4)O_(6) = 473.44`
Weight of `P_(4)O_(6)` left unreacted `(437.44 xx 0.220)/(8) = 12.03 g`


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