The bad smell of water containing hydrogen sulfide
can be removed by the action of chlorine. The reaction
is
H2S(aq) + Cl2(aq)-->2HCl(aq) + S(s)
If the hydrogen sulfide content of contaminated water
is 22 ppm by mass, calculate the amount of Cl2
(in grams) required to remove all the H2S from
2.0 . 10 2+ gallons of water. (1 gallon = 3.785 L.)
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Convert 2.0E2 gallons to grams. Convert 22 ppm to grams S. Then follow the steps in this link.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To calculate the amount of Cl2 required to remove all the H2S from the contaminated water, we need to use the given information to determine the stoichiometry of the reaction.
Step 1: Convert gallons to liters:
2.0 x 10^2 gallons x 3.785 L/gallon = 757 L
Step 2: Calculate the moles of H2S:
22 ppm (parts per million) means 22 g of H2S in 10^6 g of water.
So, the amount of H2S in 757 L of water is:
22 g/10^6 g x 757 L = 16.65 g
Step 3: Convert moles of H2S to moles of Cl2:
From the balanced equation, we can see that the stoichiometric ratio between H2S and Cl2 is 1:1.
Therefore, the moles of Cl2 required to react with the moles of H2S is also 16.65 g.
Step 4: Convert moles of Cl2 to grams:
The molar mass of Cl2 is 70.906 g/mol.
So, the mass of Cl2 required is:
16.65 g x (1 mol Cl2/70.906 g) = 0.2343 g
Therefore, approximately 0.2343 grams of Cl2 is required to remove all the H2S from 2.0 x 10^2 gallons of water.