Question

When NH3(g) reacts with O2(g), the products of the combustion are NO(g) and H2O(g). What volume of O2(g) is required to react with 3.00 mL of NH3(g)? Assume that all gases are at the same temperature and pressure.

Answer 1

Answer:

The volume of $O_2$ 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:-

$4NH_3+5O_2\rightarrow 4NO+6H_2O$

4 mL of $NH_3$ is required to react with 5 mL of $O_2$

Also,

1 mL of $NH_3$ is required to react with 5/4 mL of $O_2$

So,

3.00 mL of $NH_3$ is required to react with $\frac{5}{4}\times 3.00$ mL of $O_2$

The volume of $O_2$ required is:- 3.75 mL