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?

Typically, this reaction is catalyzed by an oxygen transfer reagent, #MnO_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 #11xx10^3*L#.

Given that #n=(PV)/(RT)# #=#

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

#=94.4*mol#

So, for each man, we need AT LEAST, #2/3xx94.4*molxx122.55*g*mol^-1=7.71xx10^3*g#, call it #8*kg# 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*kg# 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.