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2 votes
We wish to determine the

moles of lead (II) iodide
precipitated when 125 mL of
0.20 M potassium
iodide reacts with excess
lead (II) nitrate.

2Kl(aq) + Pb(NO3)2(aq) → 2KNO3(aq) + Pbl₂(s)

How many moles of KI are
present in 125 mL of 0.20 M KI?

[?] mol KI

User Jens Bodal
by
8.4k points

1 Answer

7 votes

Answer:0.025

Step-by-step explanation:

125/1000 x 0.2=0.025

User MWillis
by
8.4k points