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

Pb
(
NO
3
)
2
(
aq
)
+
2
NH
4
I
(
aq
)
⟶
PbI
2
(
s
)
+
2
NH
4
NO
3
(
aq
)

What volume of a
0.550
M NH4I solution is required to react with
751
mL of a
0.380
M Pb(NO3)2 solution?

volume:
mL

How many moles of PbI2 are formed from this reaction?

Respuesta :

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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