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")`

Answer

`(a)10"mol", (b)L, (c ) 2.5`

Explanation

`(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`