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# Chem assignment help! I did something wrong and I could really use some explanation as to how to do this question?

For the equation 4Cl2 + H2S + 4H2O —> H2SO4 + 8HCl, if 20,000L of contaminated water was in a tank with a concentration of hydrogen Sulfide of 7.13 x10-4, what volume of chlorine at STP is required to remove all hydrogen sulfide from the water?

### 4 Answers

- Roger the MoleLv 710 months ago
Supposing the missing units on "7.13 x10-4" to be "mol/L":

(20000 L) x (7.13 x10^-4 mol H2S/L) x (4 mol Cl2 / 1 mol H2S) = 57.04 mol Cl2

V = nRT / P = (57.04 mol Cl2) x (22.414 L/mol) = 1278 L Cl2

- billrussell42Lv 710 months ago
4Cl₂ + H₂S + 4H₂O —> H₂SO₄ + 8HCl

7.13 x10-4 x 20,000 L = 14.26 L of H₂S

from the balanced equation, 4 mol Cl₂ is required for 1 mol of H₂S

since both are at STP, mole ratio translates to volume ratio

therefore 14.26 L x 4 = 57.04 L of Cl₂ is required

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- Bobby_ThinLv 710 months ago
4 Cl2 + H2S + 4 H2O = H2SO4 + 8 HCl

lets assume the [ H2S] is 7.13 x10-4 moles per liter

in 20,000 L we have ,14.26 moles of H2S

according to the equation these react with 4 *14.26 moles of Cl2 = 57.0 moles

at STP 1 mole of Cl2 occupies 22,4 liters so we have 57.0 moles * 22.4 liter / mol = 1280 liters

so we need 1280 liters of Cl2 to remove the H2S

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- Anonymous10 months ago
I don't see any attempt at answering so 🤷♂️

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Probably. You can do it yourself.