`NaIO_3` reacts with `NaHSO_3` according to equation: `IO_3^(-) + 3HSO_3^(-) to I^(-) + 3H^(+) + 3SO_4^(2-)`. The weight of `NaHSO_3` required to react with 100 mL of solution containing 0.66 g of `NaIO_3` is

Options

A. 5.2g
B. 4.57g
C. 2.3 g
D. 1.04 g

Answer

D

Explanation

In the reaction, `I^(5+) + 6e^(-) to I^(-) , S^(6+) + 2e^(-) to S^(4+)`
therefore, `M_("equiv") NaHSO_3 = M_("equiv")NaIO_3`
`N_1V_1 (NaHSO_3) = N_2V_2(NaIO_3), W/(104//2) xx 100 = (0.66)/(198//6) xx 100 , w = 1.04 g`