Iodine is prepared both in the laboratory and commercially by adding Cl 2 ( g ) Cl2(g) to an aqueous solution containing sodium iodide. 2 NaI ( aq ) + Cl 2 ( g ) ⟶ I 2 ( s ) + 2 NaCl ( aq ) 2NaI(aq)+Cl2(g)⟶I2(s)+2NaCl(aq) How many grams of sodium iodide, NaI , NaI, must be used to produce 83.9 g 83.9 g of iodine, I 2 ?

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iimuetinyan iimuetinyan · Answer 1 · 2019-12-06T23:05:59+01:00

Answer:

Explanation:

The reaction of Chlorine with Sodium Iodide is given as;

2NaI + Cl2 - I2 + 2NaCl

The molar mass of iodine gas is 253.8089g/mol.

The molar mass of Sodium Iodide is 149.89g/mol.

The number of moles of I2 can be calculated as shown below;

moles of I2 = Given mass (g)/molecular mass (g/mol).

= 83.9g/253.8089g/mol.

= 0.330mol

In the above reaction, 2mol of NaI reacts with 1 mol of Cl to form 1 mol of I2 and 2 moles of NaCl.

The number of moles of NaI would be;

Number of moles of NaI = moles of I2 * 2 mol of NaI/1 mol of I2

= 0.330mol * 2/1

= 0.660mol

Therefore, the mass of NaI needed would be;

Mass of NaI = moles of NaI * molar mass of NaI

= 0.660mol * 149.89g/mol

= 98.92g

The mass of Sodium Iodide required is 98.92g.