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(This question is answered below by assuming that the lower-case "m" in the first line of the question was intended to be a capital "M" as in the second line. If the "m" was intended to mean "molality" rather than molarity, the question can not be answered without additional information, specifically the density of the Pb(NO3) solution at the temperature at which the volume specified was measured.)

The formula for a "mole" [more strictly, "formula unit"] of the lead nitrate solution given in the question shows that each unit contains one lead cation. The KI solution is stated to be in excess, so that every lead cation supplied to the reaction will be precipitated.

A 2.0 molar solution by definition contains 2.0 moles of lead cations in each liter of solution. The solution is homogeneous. therefore, 500ml of it will contain half as many lead cations, in this instance 1.0. Therefore, 1.0 moles of lead (II) iodide will be precipitated.

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Q: When 500 ml of 2.0m solution of Pb NO3 2 is mixed with an excess of 1.0 M solution of KI how many moles of lead II iodide are formed?
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When 0.500 l of a 1.00 m solution of pb no3 2 is mixed with an excess of 1.00 m solution of ki how many grams of lead II iodide are formed?

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