What weight of nitrate ion (calculated as HNO3) is needed to convert 5 g of iodine into iodic acid according to the reaction I2 + HNO3 → HIO3 + NO2 + H2O

Question

What weight of nitrate ion (calculated as HNO3) is needed to convert 5 g of iodine into iodic acid according to the reaction I2 + HNO3 → HIO3 + NO2 + H2O (A) 12.205 g (B) 24.8 g (C) 0.248 g (D) 49.6 g

Tardigrade Admin · Accepted Answer

From the given reaction, the balanced equation is

I_{2} + 10 N O_{3}^{-} + 8 H^{+} \to 2 I O_{3}^{-} + 10 N O_{2} + 4 H_{2} O

From the above equation,

254 g

of

I_{2}

requires

620 g

of

N O_{3}^{-}

ion

∴ 5 g o f I_{2}

requires

620

of

N O_{3}^{-}

ion

= \frac{620 \times 5}{254}

= 12.205

g of

N O_{3}^{-}

Q. What weight of nitrate ion (calculated as $H N O_{3}$ ) is needed to convert 5 g of iodine into iodic acid according to the reaction
$I_{2} + H N O_{3} \to H I O_{3} + N O_{2} + H_{2} O$

$12.205$ g

$24.8$ g

$0.248$ g

$49.6$ g

Q. What weight of nitrate ion (calculated as HNO3​) is needed to convert 5 g of iodine into iodic acid according to the reaction I2​+HNO3​→HIO3​+NO2​+H2​O

12.205 g

24.8 g

0.248 g

49.6 g

From the given reaction, the balanced equation is I2​+10NO3−​+8H+→2IO3−​+10NO2​+4H2​O From the above equation,254g of I2​ requires 620g of NO3−​ ion ∴5gofI2​ requires 620 of NO3−​ ion=254620×5​ =12.205 g of NO3−​

Q. What weight of nitrate ion (calculated as $H N O_{3}$ ) is needed to convert 5 g of iodine into iodic acid according to the reaction
$I_{2} + H N O_{3} \to H I O_{3} + N O_{2} + H_{2} O$

$12.205$ g

$24.8$ g

$0.248$ g

$49.6$ g