# 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.