how many moles of PbI2 are formed from this reaction? i keep getting the answer wrong ??

lead(ii) nitrate and ammonium iodide react to form lead (ii) iodide ammonium nitrate according to the reaction: Pn(NO3)2 (aq) +2NH4I(aq)—> PbI2(s)+2NH4NO3(aq) 1. what volume of 0.390 MNH4I solution is required to react with 377 mL of a 0.180 M(molarity) Pb(NO3)2 solution? 2. how many moles of PbI2 are formed from this reaction? I already figured out number 1. i just can figure out 2. I keep getting .06786 … and rounding it but it keeps saying its wrong. Please help

1 Answer

  • Unles you are rounding incorrectly (to 3 sig figs) you are correct and the given answer is wrong 2. moles PbI = 0.06786 mol Pb(NO3)2 x (1 mol PbI2 / 1 mol Pb(NO3)2) = 0.06786 mol

    = 0.0679 mol (3 sig figs)

Leave a Comment