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Alchen [17]
4 years ago
10

Lead iodide is virtually insoluble in water so that the reaction appears to go to completion. How many milliliters of 3.550 M HI

(aq) must be added to a solution containing 0.700 mol of to completely precipitate the lead?

Chemistry
2 answers:
Minchanka [31]4 years ago
8 0

Answer:

394mL

Explanation:

the molecular equation is

Pb(NO3)2 + 2 HI = PbI2 + 2 HNO3

the ratio between Pb(NO3)2 and HI is 1 : 2

moles HI required = 0.700 x 2 = 1.40

volume HI = 1.40 mol / 3.550 mol/L=0.394 L = 394 mL

Bad White [126]4 years ago
4 0

Explanation:

Below is an attachment containing the solution.

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