Answer:
The volume of
required is:- 3.75 mL
Step-by-step 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:-
![4NH_3+5O_2\rightarrow 4NO+6H_2O](https://img.qammunity.org/2020/formulas/chemistry/college/8lcg9y57i5pww9ncpkyj2hr5nlhtzefwla.png)
4 mL of
is required to react with 5 mL of
![O_2](https://img.qammunity.org/2020/formulas/chemistry/middle-school/9c0d0q54qoy7o2wh2yi3hpprbbhzj1k8rs.png)
Also,
1 mL of
is required to react with 5/4 mL of
![O_2](https://img.qammunity.org/2020/formulas/chemistry/middle-school/9c0d0q54qoy7o2wh2yi3hpprbbhzj1k8rs.png)
So,
3.00 mL of
is required to react with
mL of
![O_2](https://img.qammunity.org/2020/formulas/chemistry/middle-school/9c0d0q54qoy7o2wh2yi3hpprbbhzj1k8rs.png)
The volume of
required is:- 3.75 mL