1.

What weight of `NaHSO_3` is required to react with 100 " mL of " solution containing 0.33 g of `NaIO_3` according to the following reaction: `IO_3^(ɵ)+HSO_3^(ɵ)toI^(ɵ)+SO_3^(2-)` (a). `0.52g` (b). 5.2 g ltbr) (c). `1.04g` (d). `10.4g`

Answer» `Mw(NaHSO_3)=23+1+32+3xx16=104g`
`Mw(NaIO_3)=23+127+2xx16=198g`
`[6e^(-)+IO_3^(ɵ)toI^(ɵ)](n=6)`
`[HSO_3^(ɵ)toI^(ɵ)](n=2)`
Strength `(gL^(-1))` `=`Normality`xxEw`
Normality of `NaIO_3=("Strength")/(Ew)`
`=(3.3)/((198)/(6))=(3.3xx6)/(198)=0.100`
Strength of `NaIO_3=(0.33g)/(100mL)`
`-=(0.33xx1000)/(100)=3.3gL^(-1)`
Normality of `NaIO_3=0.1N`
m" Eq of "`NaIO_3=NxxV(mL)=0.1xx100=10mEq`
`NaHSO_3-=NaIO_3`
`mEq-=mEq-=10mEq`
`m" Eq of "NaHSO_3=10mEq`
`=10xx10^(-3)Eq`
`=10xx10^(-3)xxEw(NaHSO_3)`
`=10xx10^(-3)xx(104)/(2)g=0.52g`


Discussion

No Comment Found

Related InterviewSolutions