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?
+5
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Home » Chemistry » 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.