7) What is the mass of insoluble lead(II) iodide (461.0 g/mol) produced from 0.830 g of potassium iodide (166.00 g/mol) and aqueous lead(II) nitrate? __Pb(NO3)2(aq) + __KI(s) -> __PbI2(s) + 2KNO3(aq) A) 4.61 g B) 0.149 g C) 2.31 g D) 1.15 g E) 0.598 g
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Given: Mass of KI = 0.830 g Molar mass of KI = 166.00 g/mol Number of moles of KI = Mass/Molar mass Number of moles of KI = 0.830 g / 166.00 g/mol Number of moles of KI = 0.00500 mol Show more…
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