Iodine is prepared both in the laboratory and commercially by adding Cl2(g) to an aqueous solution containing sodium iodide: 2NaI(aq) CI2(g) --------> I2(s) 2NaCI(aq)
_____________ g NaI ?
How many grams of sodium iodide, NaI, must be used to produce 71.1 g of iodine, I2?
2017-01-13 7:55 pm
回答 (2)
2017-01-13 9:15 pm
Molar mass of NaI = (23.0 + 126.9) g/mol = 149.9 g/mol
Molar mass of I₂ = 126.9 × 2 g/mol = 253.8 g/mol
2NaI(aq) + Cl₂(g) → I₂(aq) + 2NaCl(aq)
OR: Mole ratio NaI : I₂ = 2 : 1
No. of moles of I₂ produced = (71.1 g) / (253.8 g/mol) = 0.2801 mol
No. of moles of NaI must be added = (0.2801 mol) × 2 = 0.5602 mol
Mass of NaI must be added = (0.5602 mol) × (149.9 g/mol) = 84.0 g
Molar mass of I₂ = 126.9 × 2 g/mol = 253.8 g/mol
2NaI(aq) + Cl₂(g) → I₂(aq) + 2NaCl(aq)
OR: Mole ratio NaI : I₂ = 2 : 1
No. of moles of I₂ produced = (71.1 g) / (253.8 g/mol) = 0.2801 mol
No. of moles of NaI must be added = (0.2801 mol) × 2 = 0.5602 mol
Mass of NaI must be added = (0.5602 mol) × (149.9 g/mol) = 84.0 g
2017-01-17 5:32 pm
(71.1 g I2) / (253.8089 g I2/mol) x (2 mol NaI / 1 mol I2) x (149.8942 g NaI/mol) = 84.0 g NaI
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