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When .5 L of a 1.0 m solution of Pb[NO3]2 is mixed with .5 L of a 1.0 M solution of KI, how many moles of....?

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lead II iodide are formed.

.25 mol PbI2

.5 mol PbI2

1 mol PbI2

2 mol PbI2

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  1. 0.5 L of a 1.0 m solution of Pb[NO3]2 = 0.5 moles lead

    is mixed with .5 L of a 1.0 M solution of KI = 0.5 moles Iodide

    1 Pb +2   &  2 I- --> PbI2

    2 moles iodide make 1 mol PbI2

    your answer:  .25 mol PbI2

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