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Lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide: Pb2+ (aq) + 2I− (aq) → PbI2 (s) 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 0.600 mol of Pb(NO3)2(aq) to completely precipitate the lead?

Respuesta :

IthaloAbreu

Answer:

338.00 mL

Explanation:

The lead ions come from the salt Pb(NO₃)₂ and the iodide from the acid HI, so the balanced reaction is:

Pb(NO₃)₂(aq) + 2HI(aq) → PbI₂(s) + 2HNO₃(aq)

So, the stoichiometry is 1 mol of Pb(NO₃)₂ to 2 moles of HI, then:

1 mol of Pb(NO₃)------------------------------------ 2 moles of HI

0.600 mol of Pb(NO₃)₂--------------------------- x

By a simple direct three rule:

x = 1.200 mol of HI

The acid has concentration equal to 3.550 mol/L, the volume (V) is the number of moles divided by the molar concentration:

V = 1.200/3.550 = 0.338 L

V = 338.00 mL

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