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