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Ammonia can be oxidized by oxygen to produce $NO_2$ according to reaction $4NH_3+7O_2\rightarrow 4NO_2+6H_2O$.If water is formed of $36\;mol\;L^{-1}min^{-1}$.Then respective rates of $NH_3,O_2$ used will be(in M/min)

$\begin{array}{1 1}(a)\;42,24\\(b)\;24,42\\(c)\;36,24\\(d)\;28,42\end{array}$

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Answer 24, 42
Rate = $\large\frac{d[H_2O]}{6dt}=\frac{d[NO_2]}{4dt}=-\frac{d[NH_3]}{4dt}=-\frac{d[O_2]}{7dt}$
Since, the rate for water is given at $36\;mol\;L^{-1}min^{-1}$, we can find the respective rates for $NH_3$ and $O_2$ as 24 and 42.
answered Dec 17, 2013 by sreemathi.v
edited Jul 27, 2014 by balaji.thirumalai
 

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