What mass of $H N O_{3}$ is needed to convert 5 g of iodine into iodic acid according to the reaction? (at mass of I = 127 u)

12.4 g

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24.8 g

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0.248 g

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49.6 g

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Solution

The correct option is A
12.4 g

$M_{e q} . o f H N O_{3} = M_{e q} . o f I_{2} \frac{w}{\frac{63}{1}} \times 1000 = \frac{5}{\frac{254}{10}} \times 1000 w = 12.4 g$

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Similar questions

H N O_{3}

5 g

I_{2} + H N O_{3} ⟶ H I O_{3} + N O_{2} + H_{2} O

H N O_{3}

I_{2} + H N O_{3} \to H P O_{3} + N O_{2} + H_{2} O

I_{2}

68 %

H N O_{3}

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H N O_{3}

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