# Question #ebf3f

Oct 6, 2016

It is a fact (i) that most elemental gases are DIATOMIC; it is a fact (ii) that mass is always conserved in a chemical reaction.

#### Explanation:

Most of the elemental gases, ${O}_{2}$, ${N}_{2}$, ${F}_{2}$, $C {l}_{2}$ are diatomic under standard conditions. The exceptions are the Noble Gases, which rarely form compounds. You simply have to know the so-called molecularity of oxygen, nitrogen, etc. Most inorganic chemists refer to these gases as $\text{dinitrogen}$, $\text{dioxygen}$, $\text{dihydrogen}$ to avoid ambiguity.

Knowledge of the molecularity of the elemental gases would be assumed for an A level student, and it certainly would be expected of a 1st year college student.

Now chemical reactions follow experimental result; experimental result does not follow our depiction of the equation. For the reaction of oxygen and hydrogen, we know that 1/2 an equiv of dioxygen reacts with one equiv of dihydrogen under normal conditions to give one equiv of water:

${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow {H}_{2} O \left(l\right)$ $\left(i\right)$

If I wanted to, I could double the entire equation to get rid of the half-integral coefficients.

$2 {H}_{2} \left(g\right) + {O}_{2} \left(g\right) \rightarrow 2 {H}_{2} O \left(l\right)$ $\left(i i\right)$

The point is that these equations are representations of chemical reality. Now these reactions certainly conserve mass: in $\left(i\right)$ 18 g of dioxygen and dihydrogen combine to give 18 g of water (we would assume it to be a liquid under standard conditions!).

Given certain conditions, however, we could in fact produce hydrogen peroxide under so-called reducing conditions:

${H}_{2} \left(g\right) + {O}_{2} \left(g\right) \rightarrow H O - O H \left(l\right)$ $\left(i i i\right)$

This would be admittedly hard to do, nevertheless, we can still conceive of the reaction as a formality, and we can certainly estimate the thermodynamic parameters involved in the reaction. In the reaction $\left(i i i\right)$, less energy would be evolved in the reaction because we are not completely reducing the oxygen gas, and we are only forming $2 \times H - O$ bonds rather than the $4 \times H - O$ bonds we form in $\left(i i\right)$. For another question dealing with the conservation of mass, see here.

If there is a more specific question, or query, ask away, and someone will try to help.