2NaI(aq)+Cl₂(g)-->I₂(s)+2NsCl(aq) How many grams of sodium iodine, NaI, must be used to produce 14.9 g of iodine, I₂?
回答 (1)
2017-10-16 5:56 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(s)+ 2NaCl(aq)
Mole ratio NaI : I₂ = 2 : 1
No. of moles of I₂ produced = (14.9 g) / (253.8 g/mol) = 0.0587 mol
No. of moles of NaI used = (0.0587 mol) × 2 = 0.1174 mol
Mass of NaI used = (0.1174 mol) × (149.9 g/mol) = 17.6 g
Molar mass of I₂ = 126.9 × 2 g/mol = 253.8 g/mol
2NaI(aq) + Cl₂(g) → I₂(s)+ 2NaCl(aq)
Mole ratio NaI : I₂ = 2 : 1
No. of moles of I₂ produced = (14.9 g) / (253.8 g/mol) = 0.0587 mol
No. of moles of NaI used = (0.0587 mol) × 2 = 0.1174 mol
Mass of NaI used = (0.1174 mol) × (149.9 g/mol) = 17.6 g
收錄日期: 2021-04-24 01:11:14
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