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Question: Calculate the maximum amount of \(NaO{I_3} \) that can be produced by 12.8 g of \(HI \) ? \( 6Hl{...

Calculate the maximum amount of NaOI3NaO{I_3} that can be produced by 12.8 g of HIHI ?
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 \\\

Explanation

Solution

Hint : Sodium iodate ( NaIO3NaIO_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: HIO3+NaOHNaIO3+H2OHI{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: 3I2+6NaOHNaIO3+5NaI+3H2O3{I_2} + 6NaOH \to NaI{O_3} + 5NaI + 3{H_2}O 6Hl + 2HNO3  3l2 + 2NO + 4H2O6Hl{\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 12\dfrac{1}{2} mole 3l2 + 6NaOH  NalO3 + 5Nal + 3H2O3{l_2}{\text{ }} + {\text{ }}6NaOH{\text{ }} \to {\text{ }}Nal{O_3}{\text{ }} + {\text{ }}5Nal{\text{ }} + {\text{ }}3{H_2}O 12\dfrac{1}{2} mole \to   12×13\;\dfrac{1}{2} \times \dfrac{1}{3} mole =16= \dfrac{1}{6} mole For 1 mole of HI 16\dfrac{1}{6} mole of NaOI3NaO{I_3}
Mole of HIHI =12.8128=0.1= \dfrac{{12.8}}{{128}} = 0.1 mole
Mole of NaOI3=0.1×16NaO{I_3} = 0.1 \times \dfrac{1}{6}
Mass of NaOI3=0.1×16×198NaO{I_3} = 0.1 \times \dfrac{1}{6} \times 198 3.3\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.