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?

Question

Niko Griffith

Chemistry

2 July, 01:35

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

Answers (1)

Know the Answer?

Hayes · Answer 1

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