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Water contaminated with `H_(2)S` can be freed from `H_(2)S` by passing `Cl_(2)` through it. If the `H_(2)S` content in contaminated water is `22` ppm by mass how much `Cl_(2)` is needed to remove all the `H_(2)S` from `2xx10^(2)` gallons of water. `(1 gallon =3.785 litre)` |
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Answer» `because 10^(6)g` or `mL H_(2)O` contain `22g H_(2)S` `:. 2xx 10^(2) xx 3.785 xx 10^(3) xx 10^(3) mL` `= (22 xx 2 xx 10^(2) xx 3.785 xx 10^(3))/(10^(6))` `= 22 xx 2 xx 3.785 xx 10^(-1)` `= 16.654gH_(2)S` `Cl_(2)(g)+H_(2)S(aq).rarr2HCl_(aq)+S(s)` Eq. of `Cl_(2) =` Eq. of `H_(2)S` `(w)/(35.5) = (16.654)/(34//2)` `w = 34.78 g Cl` |
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