Q: NaIO_{3} reacts with NaHSO_{3} according to equation

IO_{3}^{–} + 3HSO_{3}^{–} → I^{–} + 3H^{+}+ 3SO_{4}^{2–}

The weight of NaHSO_{3} required to react with 100 ml of solution containing 0.58 gm of NaIO_{3} is

(A) 5.2 gm

(B) 4.57 gm

(C) 2.3 gm

(D) None of these

Sol: $\large \frac{W_{NaHSO_3}}{M_{NaHSO_3}} \times 2 \times 1000= \frac{0.58}{198}\times 6 \times 100 $

$\large W_{NaHSO_3} = 4.57 gm$

Ans: (B)