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?
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
