What quantity of dihydrogen and dioxygen gas will result if a 6.2*mol of water is decomposed?

Apr 22, 2017

Well, you need a stoichiometric equation to represent the decomposition of water..........

Explanation:

And here is one such:

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

Is this stoichiometrically balanced? If it is not, then we cannot accept this as a true representation of physical reality. This is in fact a redox equation, because while the oxygen in water is formally oxidized to oxygen gas, the hydrogen in water is formally reduced to dihydrogen gas. And clearly, we would expect this process to be quite endothermic. Why so?

So you have $6.2 \cdot m o l$ of water to decompose........The stoichiometry quite clearly dictates that $6.2 \cdot m o l$ of dihydrogen gas will result, along with $3.1 \cdot m o l$ of dioxygen gas. Are you with me?

And if the gases are measured under the same conditions of temperature and pressure (and we would anticipate such measurement!), we could also specify that $\text{TWO VOLUMES}$ of dihydrogen are evolved for $\text{EACH VOLUME}$ of dioxygen gas.