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In a water treatment plant, `Cl_(2)` used for the treatment of water is produced from the following reaction `2KMnO_(4)+16HClrarr2KCl+2MnCl_(2)+8H_(2)O+5Cl_(2)`. If during each feed `1 L KMnO_(4)` having `79%(w//v) KMnO_(4) & 9 L HCl` with `d=1.825 g//mL & 10% (w//w) HCl `are entered `&` if that percent yield is `80%` then calculate : `(a)` amount of `Cl_(2)` produced. `(b)` amount of water that can be treated by `Cl_(2)` if `1` litre consumes `28.4 g Cl_(2)` for treatment, `(c)` calculate efficiency `eta` of the process is `eta = ("vol. of water treated")/("vol. of total feed")` |
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Answer» Correct Answer - `(a)10"mol", (b)L, (c ) 2.5` `(a)1LKMnO_(4)rarr 79%(w//v)i.e.100mL` solution contain `79 g KMnO_(4)` moles of `KMnO_(4)=(et)/(M_(w))=(79)/(158)=0.5` Molarity `(M) = (0.5)/(100)xx1000=5M` `HClrarr10%(w//w)i.e.100g` solution contain `10g HCl` `D=1.825g//mL` `V=(M)/(D)=(100)/(1.825xx1000)` Molarity `=(10xx1.825xx1000)/(36.5xx100)=5M` `2KMnO_(4)+16HClrarr2KCl+2MnCl_(2)+8H_(2)O+5Cl_(2)` `MxxV_(1)" "MxxV_(1)` `5xx1" "5xx9` `5" "5` `-" "5" "12.5` `Cl_(2)=12.5xx(80)/(100)=10`mol. (b) `2KMnO_(4)+16HClrarr2KCl+2MnCl_(2)+8H_(2)O+5Cl_(2)` `1xx(710)/(28.4)=25L` (c) `eta = ("vol of water treated")/("vol of" KMnO_(4)+HCl) = (25)/(1+9)=2.5` |
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