To obtain maximum mass of `NO_(2)` from a given mass of a mxiture of `NH_(3)` and `O_(2)`, the ratio of mass of `NH_(3)` to `O_(2)` should be `2NH_(3)+(7)/(2)O_(2) rarr 2NO_(2)+3H_(2)O`
A
`(17)/(40)`
B
`(4)/(7)`
C
`(17)/(56)`
D
None of these
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For the given series of reaction, 4NH_(3)(g) + 5O_(2)(g) rarr 4NO(g) + 6H_(2)O(l) 2NO(g) + O_(2)(g) rarr 2NO(g) +O_(2)(g) rarr 2NO_(2)(g) To obtain maximum mass of NO_(2) from a given mass of a mixture of NH_(3) and O_(2) the ratio of mass of NH_(3) ot O_(2) should be:
NH_(3) and H_(2)O form NH_(4)OH by
NH_(4)NO_(3) overset(Delta) to N_(2)O+H_(2)O
Mass ratio of NH_(3) and (CO_(2) for maximum product formation as per reaction : 2NH_(3) + CO_(2) rArr NHCOONH_(4)
Equivalent masses of NH_(3) in the reactions are i) 4NH_(3)+5O_(2)rarr4NO+6H_(2)O ii) 2NH_(3)rarr N_(2)+3H_(2) (A) 5:6 (B) 6:5 (C) 5:3 (D) 3:5
The IUPAC name of K_(2)[Cr(CN)_(2)O_(2)(O_(2))]NH_(3)
