Iodine is prepared both in the laboratory and commercially by adding cl2 (g) to an aqueous solution containing sodium iodide: how many grams of sodium iodide, nai, must be used to produce 48.7 g of iodine, i2?

Question

Chemistry

Leilani Obrien

1 August, 19:51

Iodine is prepared both in the laboratory and commercially by adding cl2 (g) to an aqueous solution containing sodium iodide: how many grams of sodium iodide, nai, must be used to produce 48.7 g of iodine, i2?

+4

Answers (1)

Know the Answer?

Octavio Coffey · Answer 1

The balanced equation for the reaction between NaI and Cl₂

2NaI and Cl₂ - - > 2NaCl + I₂

stoichiometry of NaI to I₂ is 2:1

this means that when 2 mol of NaI reacts, it produces 1 mol of I₂

Molar mass of I₂ is 254 g/mol

Number of I₂ moles produced - 48.7 g / 254 g/mol = 0.192 mol

for 1 mol of I₂ to be formed - 2 mol of NaI required

Therefore to form 0.192 mol of I₂ - 2 x 0.192 = 0.384 mol of NaI

Then mass of NaI required - 0.384 mol x 150 g/mol = 57.6 g