
Calculate the maximum amount of $NaO{I_3} $ that can be produced by 12.8 g of $HI $ ?
$ 6Hl{\text{ }} + {\text{ }}2HN{O_3}{\text{ }} \to {\text{ }}3{l_2}{\text{ }} + {\text{ }}2NO{\text{ }} + {\text{ }}4{H_2}O \\
3{l_2}{\text{ }} + {\text{ }}6NaOH{\text{ }} \to {\text{ }}Nal{O_3}{\text{ }} + {\text{ }}5Nal{\text{ }} + {\text{ }}3{H_2}O \\ $
Answer
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Hint :Sodium iodate ( $NaIO_3 $) is the sodium salt of iodic acid. Sodium iodate is an oxidizing agent, and as such it can cause fires upon contact with combustible materials or reducing agents.Sodium iodate is the sodium salt of iodic acid. It is an oxidizing agent and dough conditioner used to strengthen doughs. It is commonly used in leavened products such as bread, rolls, and sweet rolls. It is commonly found as a white crystalline powder and has the following physical properties:White, crystalline powder.Water and acetic acid soluble
Complete Step By Step Answer:
It can be prepared by reacting a sodium-containing base such as sodium hydroxide with iodic acid, for example: $HI{O_3} + NaOH \to NaI{O_3} + H2O $ It can also be prepared by adding iodine to a hot, concentrated solution of sodium hydroxide or sodium carbonate: $3{I_2} + 6NaOH \to NaI{O_3} + 5NaI + 3{H_2}O $ $6Hl{\text{ }} + {\text{ }}2HN{O_3}{\text{ }} \to {\text{ }}3{l_2}{\text{ }} + {\text{ }}2NO{\text{ }} + {\text{ }}4{H_2}O $
In the above reaction 1 mole of $ \to $ $\dfrac{1}{2} $ mole $3{l_2}{\text{ }} + {\text{ }}6NaOH{\text{ }} \to {\text{ }}Nal{O_3}{\text{ }} + {\text{ }}5Nal{\text{ }} + {\text{ }}3{H_2}O $ $\dfrac{1}{2} $ mole $ \to $ $\;\dfrac{1}{2} \times \dfrac{1}{3} $ mole $ = \dfrac{1}{6} $ mole For 1 mole of HI $\dfrac{1}{6} $ mole of $NaO{I_3} $
Mole of $HI $ $ = \dfrac{{12.8}}{{128}} = 0.1 $ mole
Mole of $NaO{I_3} = 0.1 \times \dfrac{1}{6} $
Mass of $NaO{I_3} = 0.1 \times \dfrac{1}{6} \times 198 $ $ \to 3.3 $ gm.
Note :
Conditions/substances to avoid are: heat, shock, friction, combustible materials, reducing materials, aluminium, organic compounds, carbon, hydrogen peroxide, sulfides. The main use of sodium iodate in everyday life is in iodised salt. The other compounds which are used in iodised table salt are potassium iodate, potassium iodide, and sodium iodide. Sodium iodate comprises 15 to 50 mg per kilogram of applicable salt.
Complete Step By Step Answer:
It can be prepared by reacting a sodium-containing base such as sodium hydroxide with iodic acid, for example: $HI{O_3} + NaOH \to NaI{O_3} + H2O $ It can also be prepared by adding iodine to a hot, concentrated solution of sodium hydroxide or sodium carbonate: $3{I_2} + 6NaOH \to NaI{O_3} + 5NaI + 3{H_2}O $ $6Hl{\text{ }} + {\text{ }}2HN{O_3}{\text{ }} \to {\text{ }}3{l_2}{\text{ }} + {\text{ }}2NO{\text{ }} + {\text{ }}4{H_2}O $
In the above reaction 1 mole of $ \to $ $\dfrac{1}{2} $ mole $3{l_2}{\text{ }} + {\text{ }}6NaOH{\text{ }} \to {\text{ }}Nal{O_3}{\text{ }} + {\text{ }}5Nal{\text{ }} + {\text{ }}3{H_2}O $ $\dfrac{1}{2} $ mole $ \to $ $\;\dfrac{1}{2} \times \dfrac{1}{3} $ mole $ = \dfrac{1}{6} $ mole For 1 mole of HI $\dfrac{1}{6} $ mole of $NaO{I_3} $
Mole of $HI $ $ = \dfrac{{12.8}}{{128}} = 0.1 $ mole
Mole of $NaO{I_3} = 0.1 \times \dfrac{1}{6} $
Mass of $NaO{I_3} = 0.1 \times \dfrac{1}{6} \times 198 $ $ \to 3.3 $ gm.
Note :
Conditions/substances to avoid are: heat, shock, friction, combustible materials, reducing materials, aluminium, organic compounds, carbon, hydrogen peroxide, sulfides. The main use of sodium iodate in everyday life is in iodised salt. The other compounds which are used in iodised table salt are potassium iodate, potassium iodide, and sodium iodide. Sodium iodate comprises 15 to 50 mg per kilogram of applicable salt.
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