A 75.0 mL sample of 1.00 M sodium bromide was reacted with50.8 mL of a 0.225 M l

Question

A 75.0 mL sample of 1.00 M sodium bromide was reacted with50.8 mL of a 0.225 M lead (II) nitrate solution. a.) What type of reaction is this? b.) Give the net ionic equation for this process. c.) How much lead (II) bromide can be recovered? d.) Will any lead (II) ion be left in the solution? A 75.0 mL sample of 1.00 M sodium bromide was reacted with50.8 mL of a 0.225 M lead (II) nitrate solution. a.) What type of reaction is this? b.) Give the net ionic equation for this process. c.) How much lead (II) bromide can be recovered? d.) Will any lead (II) ion be left in the solution?

Explanation / Answer

We Know that :      2 NaBr + Pb(NO3)2 ----->  PbBr2 + 2 NaNO3  Double displacement / precipitation reaction       2 Na+ Br-    Pb+2   2NO3-   ------> PbBr2 + 2 Na+ NO3-       Net ionic equation :         Pb+2 + 2 Br-  -------> PbBr2     Number of moles of NaBr = 1.0 M x 0.075 L                                                = 0.075 moles      Number of moles of Pb(NO3)2 = 0.225 M x 0.0508L                                                        = 0.01143 moles         number of molesof PbBr2 formed is 0.01143 moles asPb(NO3)2 acts as Limiting reagent.         As the limitingreagent is Pb (NO3)2 so their is no lead ionis left unreacted.     Number of moles of NaBr = 1.0 M x 0.075 L                                                = 0.075 moles      Number of moles of Pb(NO3)2 = 0.225 M x 0.0508L                                                        = 0.01143 moles         number of molesof PbBr2 formed is 0.01143 moles asPb(NO3)2 acts as Limiting reagent.         As the limitingreagent is Pb (NO3)2 so their is no lead ionis left unreacted.

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