Chlorine dioxide `(ClO_(2))` is used now days for water treatment rather than `Cl_(2)` `ClO_(2)` is obtained by passing `(Cl_(2)(g))` into a concentrated solution of sodium chlorite `(NaClO_(2))`. The reaction gives `90%` yield. How many moles of `ClO_(2)` is produced in 3.78 L of 2.0 M `NaClO_(2)(aq)`?

Explanation

`2NaClO_(2)+Cl_(2)to2NaCl+2ClO_(2)`
Moles of `ClO_(2)=` moles of Nales of `NaClO_(2)=MxxV_(L)=2xx3.78`
`=7.56`
Moles of `ClO_(2)` actually roduced `(90%` yield) `=(7.56xx90)/(100)`
`=6.8 " mol of "ClO_(2)`