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Iodine is prepared both in the laboratory and commercially by adding Cl2(g)Cl2(g) to an aqueous solution containing sodium iodide. 2NaI(aq)+Cl2(g)⟶I2(s)+2NaCl(aq) 2NaI(aq)+Cl2(g)⟶I2(s)+2NaCl(aq) How many grams of sodium iodide, NaI,NaI, must be used to produce 89.1 g89.1 g of iodine, I2?I2? mass: g NaI

Respuesta :

Answer : The mass of sodium iodide used to produced must be, 105.22 grams.

Explanation : Given,

Mass of [tex]I_2[/tex] = 89.1 g

Molar mass of [tex]I_2[/tex] = 253.8 g/mole

Molar mass of [tex]NaI[/tex] = 149.89 g/mole

First we have to calculate the moles of [tex]I_2[/tex].

[tex]\text{Moles of }I_2=\frac{\text{Mass of }I_2}{\text{Molar mass of }I_2}=\frac{89.1g}{253.8g/mole}=0.351moles[/tex]

Now we have to calculate the moles of [tex]NaI[/tex].

The balanced chemical reaction is,

[tex]2NaI(aq)+Cl_2(g)\rightarrow I_2(s)+2NaCl(aq)[/tex]

From the balanced chemical reaction, we conclude that

As, 1 mole of [tex]I_2[/tex] obtained from 2 moles of [tex]NaI[/tex]

So, 0.351 moles of [tex]I_2[/tex] obtained from [tex]2\times 0.351=0.702[/tex] moles of [tex]NaI[/tex]

Now we have to calculate the mass of [tex]NaI[/tex].

[tex]\text{Mass of }NaI=\text{Moles of }NaI\times \text{Molar mass of }NaI[/tex]

[tex]\text{Mass of }NaI=(0.702mole)\times (149.89g/mole)=105.22g[/tex]

Therefore, the mass of sodium iodide used to produced must be, 105.22 grams.

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