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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