Answer : The limiting reagent is \(O_2\).
Solution : Given,
Mass of \(NH_3\) = 4.0 g
Mass of \(O_2\) = 8.0 g
Molar mass of \(NH_3\) = 17 g/mole
Molar mass of \(O_2\) = 32 g/mole
First we have to calculate the moles of \(NH_3\) and \(O_2\).
\(\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}=\frac{4.0g}{17g/mole}=0.24moles\)
\(\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{8.0g}{32g/mole}=0.25moles\)
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
\(4NH_3+5O_2\rightarrow 4NO+6H_2O\)
From the balanced reaction we conclude that
As, 5 mole of \(O_2\) react with 4 mole of \(NH_3\)
So, 0.25 moles of \(O_2\) react with \(\frac{0.25}{5}\times 4=0.20\) moles of \(NH_3\)
From this we conclude that, \(NH_3\) is an excess reagent because the given moles are greater than the required moles and \(O_2\) is a limiting reagent and it limits the formation of product.
Hence, the limiting reagent is \(O_2\).
