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A mixture of `H_(2(g)), N_(2(g))` and `O_(2(g))` occuping 10 ml underwent reaction so as to from `H_(2)O_(2(l))` and `N_(2)H_(2(g))` as the only products causing the volume to contacts by 6ml. The remaining mixture was passed through Pyragallol causing a contraction of 1ml. to the remaining mixture excess `H_(2)` was added and the above reaction was repeated, causing a reduction in volume of 1ml. assuming no other products to be forming calculate the volume of `H_(2)` (in ml) in the initial mixture. |
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Answer» Correct Answer - 4 `{:(H_(2(g)),,N_(2)(g),,O_(2(g))),(x+y,,x+1,,1+y),(H_(2(g)),+,O_(2(g)),to, H_(2)O_(2(l))),(x,,x,,),(H_(2(g)),+,N_(2(g)),to,N_(2)H_(2(g))),(y,,y,,y):}` 2x+2y=6 2x+2y+2=10 x+y=4 `ul(2x+y=6)` x=2 y=2 |
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