# Calculate the number of moles of precipitate that would be produced when two solutions were mixed. a) 1.0 mL 0.10 M Pb(NO3)2 and 5.0 mL, 0.10 M KIb) 2.0 mL 0.10 M Pb(NO3)2 and 4.0 mL, 0.10 M KIc) 3.0 mL 0.10 M Pb(NO3)2 and 3.0 mL, 0.10 M KId) 4.0 mL 0.10 M Pb(NO3)2 and 2.0 mL, 0.10 M KIe) 5.0 mL 0.10 M Pb(NO3)2 and 1.0 mL, 0.10 M KI

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Calculate the number of moles of precipitate that would be produced when two solutions were mixed.

a) 1.0 mL 0.10 M Pb(NO3)2 and 5.0 mL, 0.10 M KI

b) 2.0 mL 0.10 M Pb(NO3)2 and 4.0 mL, 0.10 M KI

c) 3.0 mL 0.10 M Pb(NO3)and 3.0 mL, 0.10 M KI

d) 4.0 mL 0.10 M Pb(NO3)2 and 2.0 mL, 0.10 M KI

e) 5.0 mL 0.10 M Pb(NO3)2 and 1.0 mL, 0.10 M KI

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Step 1

PbI2 is precipitated when Pb(NO3)2 and KI.

Step 2

(a)

Number of moles of precipitate (PbI2) formed is calculated as follows,

1 mL = 0.001 L

Step 3

(b)

Number of moles of precipitate (PbI2) form...

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