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`H_(2)` gas is often used as a reducing gas. In a particular set up 17.4 gm of `MnO_(2)` on reacting with excess of hydrogen gas given water & new oxide `Mn_(x)O_(y)` such that mass of the oxide obtained is 12.6. what will be value of y if x is 2.[Mn=55] |
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Answer» Correct Answer - 1 `underset(17.4 gm) (MnO_(2)) +H_(2)("excess") to H_(2)O +Mn_(2)O_(y)` POAc onn Mn `1xx17.4/87=2xx12/6/(110+16y)` `110+16y=(2xx12.6)/17.4 xx87 =126` `16y=16 rArr 1` |
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