Answer:
The volume of [tex]O_2[/tex] required is:- 3.75 mL
Explanation:
At constant pressure and temperature, the volume of the gases can be used for stoichiometric calculations in terms of moles.
Thus, For the given reaction:-
[tex]4NH_3+5O_2\rightarrow 4NO+6H_2O[/tex]
4 mL of [tex]NH_3[/tex] is required to react with 5 mL of [tex]O_2[/tex]
Also,
1 mL of [tex]NH_3[/tex] is required to react with 5/4 mL of [tex]O_2[/tex]
So,
3.00 mL of [tex]NH_3[/tex] is required to react with [tex]\frac{5}{4}\times 3.00[/tex] mL of [tex]O_2[/tex]
The volume of [tex]O_2[/tex] required is:- 3.75 mL