Question #7ea2e

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Question #7ea2e

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Answer 1 · 2016-12-22T10:11:06

The interacting ratio of number of moles of $N H_{3} and O_{2}$ $NH_3 and O_2$ as per above equation is $4 : 5 = \frac{4}{5} : 1 = 0.8 : 1$ $4:5=4/5:1=0.8:1$

So if the 1 mole of each reactant here is made to react, $O_{2}$ $O_2$ will be fully consumed reacting with $0.8$ $0.8$ mole $N H_{3}$ $NH_3$ and as a result $0.2$ $0.2$ mole of $N H_{3}$ $NH_3$ only will remain as residue.

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