Once ignited, it will produce 120 cubic feet of $O_{2}$ $O_2$ at 0.5 psi, enough to support 100 persons. After balancing the equation, how do you compute the mass of $N a C l O_{3}$ $NaClO_3$ required to generate the moles of gas needed to support the 100 crew members at 300K?

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Once ignited, it will produce 120 cubic feet of $O_{2}$ $O_2$ at 0.5 psi, enough to support 100 persons. After balancing the equation, how do you compute the mass of $N a C l O_{3}$ $NaClO_3$ 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: $N a C l O_{3} (s) \to N a C l (s) + O_{2} (g)$ $NaClO_3(s) -> NaCl(s) + O_2(g)$ .

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Answer 1 · 2016-12-17T14:13:01

$N a C l O_{3} (s) \to N a C l (s) + \frac{3}{2} O_{2} (g) ↑ ⏐ ⏐ ⏐$ $NaClO_3(s) rarr NaCl(s) + 3/2O_2(g)uarr$

Explanation:

Typically, this reaction is catalyzed by an oxygen transfer reagent, $M n O_{2}$ $MnO_2$ . 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 $11$ $11$ $m^{3}$ $m^3$ of oxygen gas per day. This is $11 \times 10^{3} \cdot L$ $11xx10^3*L$ .

Given that $n = \frac{P V}{R T}$ $n=(PV)/(RT)$ $=$ $=$

$\frac{21 % \times 1 \cdot a t m \times 11 \times 10^{3} \cdot L}{0.0821 \cdot L \cdot a t m \cdot K^{- 1} \cdot m o l^{- 1} \times 298 \cdot K}$ $(21%xx1*atmxx11xx10^3*L)/(0.0821*L*atm*K^-1*mol^-1xx298*K)$

$= 94.4 \cdot m o l$ $=94.4*mol$

So, for each man, we need AT LEAST, $\frac{2}{3} \times 94.4 \cdot m o l \times 122.55 \cdot g \cdot m o l^{- 1} = 7.71 \times 10^{3} \cdot g$ $2/3xx94.4*molxx122.55*g*mol^-1=7.71xx10^3*g$ , call it $8 \cdot k g$ $8*kg$ 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 $100 \cdot k g$ $100*kg$ 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.

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Once ignited, it will produce 120 cubic feet of $O_{2}$ $O_2$ at 0.5 psi, enough to support 100 persons. After balancing the equation, how do you compute the mass of $N a C l O_{3}$ $NaClO_3$ 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: $N a C l O_{3} (s) \to N a C l (s) + O_{2} (g)$ $NaClO_3(s) -> NaCl(s) + O_2(g)$ .

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Explanation:

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Once ignited, it will produce 120 cubic feet of O2O_2 at 0.5 psi, enough to support 100 persons. After balancing the equation, how do you compute the mass of NaClO3NaClO_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: NaClO3(s)→NaCl(s)+O2(g)NaClO_3(s) -> NaCl(s) + O_2(g).

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Explanation:

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Once ignited, it will produce 120 cubic feet of $O_{2}$ $O_2$ at 0.5 psi, enough to support 100 persons. After balancing the equation, how do you compute the mass of $N a C l O_{3}$ $NaClO_3$ 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: $N a C l O_{3} (s) \to N a C l (s) + O_{2} (g)$ $NaClO_3(s) -> NaCl(s) + O_2(g)$ .