Once ignited, it will produce 120 cubic feet of O_2 at 0.5 psi, enough to support 100 persons. After balancing the equation, how do you compute the mass of NaClO_3 required to generate the moles of gas needed to support the 100 crew members at 300K?

On submarines, chlorate candles are used to produce emergency oxygen. Once ignited, it decomposes via: $N a C l {O}_{3} \left(s\right) \to N a C l \left(s\right) + {O}_{2} \left(g\right)$.

Dec 17, 2016

$N a C l {O}_{3} \left(s\right) \rightarrow N a C l \left(s\right) + \frac{3}{2} {O}_{2} \left(g\right) \uparrow$

Explanation:

Typically, this reaction is catalyzed by an oxygen transfer reagent, $M n {O}_{2}$. This is a very convenient preparation of laboratory oxygen (and also, as you say, on board a submarine or submersible! This is not a use of which I would like to have experience.)

Your question is rather open-ended, so I will try to address it like this. From this site, we learn that the average punter consumes $11$ ${m}^{3}$ of oxygen gas per day. This is $11 \times {10}^{3} \cdot L$.

Given that $n = \frac{P V}{R T}$ $=$

(21%xx1*atmxx11xx10^3*L)/(0.0821*L*atm*K^-1*mol^-1xx298*K)

$= 94.4 \cdot m o l$

So, for each man, we need AT LEAST, $\frac{2}{3} \times 94.4 \cdot m o l \times 122.55 \cdot g \cdot m o {l}^{-} 1 = 7.71 \times {10}^{3} \cdot g$, call it $8 \cdot k g$ per day. For 100 persons, that's about a tonne of oxygen candles daily. I don't think this could be done realistically.

You are certainly free to question my assumptions, and calculations, but I suspect that oxygen candles were supplied to the sailors who manned midget submarines, i.e. with a crew of 1-2, where storage of $100 \cdot k g$ or so of sodium chlorate might be feasible. The sailors who did this, i.e. travelled with these quantities of potent oxidant into a war zone in a submersible, along with their explosives, were certainly brave and desperate men.