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Consider the balanced equation of KI reacting with Pb(NO3)2 to form a precipitate. 2KI(aq)+Pb(NO3)2(aq)⟶PbI2(s)+2KNO3(aq) What mass...

Consider the balanced equation of KI reacting with Pb(NO3)2 to form a precipitate.

2KI(aq)+Pb(NO3)2(aq)⟶PbI2(s)+2KNO3(aq)

What mass of PbI2 can be formed by adding 0.413 L of a 0.140 M solution of KI to a solution of excess Pb(NO3)2?

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Answer #1

2KI (ae) + Pb (Now Pb I2 es + 2lNaz (9) mol KIA OX au 0.140 x .413 to PoE = 109.0578 mal mol Pb 12 = £ xo.9578 = 0.0289 may m

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Consider the balanced equation of KI reacting with Pb(NO3)2 to form a precipitate. 2KI(aq)+Pb(NO3)2(aq)⟶PbI2(s)+2KNO3(aq) What mass...
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