H2S (g) concentrations of 660ppm is dangerous. (6) H2S generators continually release all of the gas they can every 15 minutes & the gas is never vented out of a room (vol 198,200L). How long will it take the H2S (g) concentration 2 reach dangerous level?

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H2S (g) concentrations of 660ppm is dangerous. (6) H2S generators continually release all of the gas they can every 15 minutes & the gas is never vented out of a room (vol 198,200L). How long will it take the H2S (g) concentration 2 reach dangerous level?

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Answer 1 · 2015-02-10T10:44:51

I don't know if the text of the question is complete, since no mention of how much gas one generator produces in 15 minutes. I'll show everything you have to do to solve this problem up until that point.

So, you know that the threshold for hydrogen sulfide ( $H_{2} S$ $H_2S$ ) is $660 ppm$ $"660 ppm"$ . Parts-per-million, or ppm, is usually used to express a very, very small concentration in solution

$1 ppm = \frac{1 g}{1000 kg} = \frac{1 g}{10^{6} g}$ $"1 ppm" = "1 g"/"1000 kg" = "1 g"/(10^(6)"g")$

For gaseous mixtures like you have here, ppmv, or parts-per-million by volume, is used instead of ppm

$1 ppmv = \frac{1 L}{10^{6} L}$ $"1 ppmv" = "1 L"/(10^(6)"L")$

Now, I assume that you must use ppm, since ppmv would require additional information, so the fastest way of doing so is by using the density of air, which is approximately ${1.20 kg/m}^{3}$ $"1.20 kg/m"^3$ at room temperature ( $20^{\circ} C$ $"20"^@"C"$ ).

Since I want to keep the calculation simple, I'll assume that the air has a density of ${1.00 kg/m}^{3}$ $"1.00 kg/m"^3$ . Since ${1 m}^{3} = 1000 L$ $"1 m"^3 = "1000 L"$ , the density of air in kg per liter will be

$1.00 \frac{kg}{m^{3}} \cdot \frac{{1 m}^{3}}{1000 L} = 10^{- 3} kg/L$ $1.00"kg"/"m"^3 * "1 m"^3/"1000 L" = 10^(-3) "kg/L"$

In this case, a concentration of 660 ppm would mean that the room must contain

$660 ppm = \frac{660 g H_{2} S}{10^{6} g of air}$ $"660 ppm" = ("660 g" H_2S)/(10^(6)"g of air")$

The volume of the room is $198,200 L$ $"198,200 L"$ , which means, using the assumption about air density made earlier, that you have

$198,200 L \cdot \frac{10^{- 3} kg}{L} \cdot \frac{10^{3} g}{1 kg} = 198,200 g$ $"198,200 L" * (10^(-3)"kg")/"L" * (10^(3)"g")/"1 kg" = "198,200 g"$ of air in the room

As a result, the mass of hydrogen sulfide will be

$\frac{660 g}{10^{6} g of air} \cdot 198,200 g of air = 130.8 g H_{2} S$ $"660 g"/(10^(6)"g of air") * "198,200 g of air" = "130.8 g"H_2S$

I assume that you have six $H_{2} S$ $H_2S$ generators, which means that each must contribute

$130.8 g \cdot \frac{1}{6} = 21.8 g$ $"130.8 g" * 1/6 = "21.8 g"$

Let's say every generator produces $X g$ $"X g"$ of $H_{2} S$ $H_2S$ every 15 minutes. The time needed to reach this quantity will be

$21.8 g \cdot \frac{15 minutes}{X g} = \frac{327}{X} minutes$ $"21.8 g" * "15 minutes"/"X g" = "327"/"X" "minutes"$

SIDE NOTE. The generator is actually called a Kipp generator, also known as KIpp's apparatus

http://www.daviddarling.info/encyclopedia/K/Kipps_apparatus.html

The apparatus uses the reaction between ferrous sulfide and a strong acid (usually hydrochloric acid) to produce hydrogen sulfide gas

$F e S + 2 H C l \to F e C l_{2} + H_{2} S$ $FeS + 2HCl -> FeCl_2 + H_2S$

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H2S (g) concentrations of 660ppm is dangerous. (6) H2S generators continually release all of the gas they can every 15 minutes & the gas is never vented out of a room (vol 198,200L). How long will it take the H2S (g) concentration 2 reach dangerous level?

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