If 3.50g of lead(II) iodide was obtained by adding excess HI to 150 ml of lead(II) nitrate, what was the molar

Question

Lead(ll)nitrate and hydroiodic acid react to produce lead(ll)iodide and nitric acid?
What was the molarity of the lead(II) nitrate solution?
If 3.50g of lead(II) iodide was obtained by adding excess HI to 150 ml of lead(II) nitrate, what was the molarity of the lead(II) nitrate solution?

Answer 1

Chemical equation:

HI + Pb(NO3)2 ~~> HNO3 + PbI2

Balanced equation:

2 HI + Pb(NO3)2 ~~> 2 HNO3 + PbI2

So, the mole ratio of Lead (II) nitrate to Lead(II) Iodide is 1:1

Molar Mass PbI2 = 207.2 + 127 * 2 = 461.2 g/mol

3.50 / 461.2 = 0.00759 mol PbI2

0.00759 mol PbI2 = 0.00759 mol Pb(NO3)2

M = mol / L

150 mL = 0.150 L

M = 0.00759 / 0.150 = 0.0506 M Pb(NO3)2

Answer 2

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