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Kasturba Biren Narain
Asked: 2022-11-06T19:27:08+05:30 In:

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`

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`
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  1. `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`

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