249 g of potassium iodide, KI, is mixed with 496.5 g of lead(II) nitrate, Pb(NO3)2. The equation of the reaction is represented below. [Pb(NO3)2 = 331; KI = 166; PbI2 = 461, KNO3 = 101] Calculate the number of moles of excess reagent left. Give your answer to three significant figures.
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Answer:
2
K
I
(
a
q
)
+
P
b
(
N
O
3
)
2
(
a
q
)
→
P
b
I
2
(
s
)
⏐
↓
+
2
K
N
O
3
(
a
q
)
.
Moles of
m
o
l
K
I
≡
1
2
⋅
m
o
l
P
b
I
2
(
s
)
Explanation:
You have done the hard yakka in quoting the balanced chemical equation. Here, the potassium iodide is clearly the limiting reagent. By the stoichiometry of the reaction, there will be half an equiv of lead iodide per equiv of potassium iodide.
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