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. 3.75 mL

Answer 1

Answer: The volume of oxygen gas required is 3.75 mL

Explanation:

STP conditions:

1 mole of a gas occupies 22.4 L of volume.

We are given:

Volume of ammonia reacted = 3.00 mL = 0.003 L    (Conversion factor:  1 L = 1000 mL)

The chemical equation for the reaction of ammonia with oxygen follows:

$4NH_3(g)+5O_2\rightarrow 4NO(g)+6H_2O(g)$

By Stoichiometry of the reaction:

(4 × 22.4) L of ammonia reacts with (5 × 22.4) L of oxygen gas

So, 0.003 L of ammonia will react with = $\frac{(5\times 22.4)}{(4\times 22.4)}\times 0.003=0.00375L=3.75mL$ of oxygen gas

Hence, the volume of oxygen gas required is 3.75 mL