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# In the reaction,$4NH_3+5O_2\rightarrow 4NO+6H_2O$ when one mole of ammonia and one mole of oxygen are made to react completely then

$\begin{array}{1 1}(a)\;\text{1.0 mole of NO is formed}\\(b)\;\text{all oxygen is consumed} \\(c)\;\text{1.0 mole of }H_2O\text{ is produced}\\(d)\;\text{all ammonia is consumed}\end{array}$

As 4 mole of $NH_3$ combine with 5 mol of $O_2$
$\therefore$ 1 mol of $NH_3$ combine with $\large\frac{5}{4}=$$1.25$mol of $CO_2$
Therefore,$O_2$ is the limiting reactant.
Hence (b) is the correct option.