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densk [106]
3 years ago
9

Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction

Chemistry
1 answer:
inessss [21]3 years ago
6 0

Answer:

Explanation:

2 NH4I    + Pb(NO3)2   =  PbI2  + 2 NH4NO3

2 mol          1 mol

x ml             751 ml

0.550 M      0.380 M

0.380 mol/1000 ml  x 751 ml =  0.285 mol Pb(NO3)2

0.285 mol Pb(NO3)2 x 2 mol NH4I/1 mol  Pb(NO3)2=0.571 mol  NH4I  needed

0.571 mol  NH4I/0.550 mol/1000 ml=1038 ml

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