Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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)

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

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App