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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?

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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
User Mrye
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