Pb(NO3)2(aq )+ 2KI(s) ---> PbI2(s) + 2KNO3(aq)
please help :( im stuck on this question. please explain how you got the answer!
whats is the mass of potassium iodide (166.00g/mol) that yields 0.500 g of lead (II) iodide (461.0g/mol) precipitate?
2017-11-06 12:07 pm
回答 (2)
2017-11-06 12:23 pm
Pb(NO₃)₂(aq ) + 2KI(s) → PbI₂(s) + 2KNO₃(aq)
Reaction of 2 moles of KI yield 1 mole of PbI₂.
No. of moles of PbI₂ produced = (0.500 g) / (461.0 g/mol) = 0.001085 mol
No. of moles of KI needed = (0.001085 mol) × 2 = 0.00217 mol
Mass of KI needed = (0.00217 mol) × (166.00 g/mol) = 0.360 g
Reaction of 2 moles of KI yield 1 mole of PbI₂.
No. of moles of PbI₂ produced = (0.500 g) / (461.0 g/mol) = 0.001085 mol
No. of moles of KI needed = (0.001085 mol) × 2 = 0.00217 mol
Mass of KI needed = (0.00217 mol) × (166.00 g/mol) = 0.360 g
2017-11-06 12:23 pm
Supposing an excess of Pb(NO3)2:
(0.500 g PbI2) / (461.0 g PbI2/mol) x (2 mol KI / 1 mol PbI2) x (166.00 g KI/mol) = 0.360 g KI
(0.500 g PbI2) / (461.0 g PbI2/mol) x (2 mol KI / 1 mol PbI2) x (166.00 g KI/mol) = 0.360 g KI
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