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0.083grams of lead nitrate, Pb(NO3)2, in solutions mixed with 0.300g of sodium iodide in solution. This results in the formation of a yellow precipitate of lead iodide. Calculate the weight of the precipitate.

  • chemistry -

    Pb(NO3)2 + 2NaI ==> PbI2 + 2NaNO3

    Convert 0.083 g Pb(NO3)2 to moles. #moles = grams/molar mass
    Convert moles Pb(NO3)2 to moles PbI2 using the coefficients in the balanced equation which I wrote.
    Convert moles PbI2 to grams. grams = moles x molar mass.

  • chemistry -

    the 0.300g is not needed?

  • chemistry -

    Yes, it IS needed but I overlooked it. However, there are 0.002 moles of NaI which is more than enough to react with all of the Pb(NO3)2 (0.00025 moles); therefore, the Pb(NO3)2 is the limiting reagent and the problem is worked exactly as I showed it. That is, the PbI2 formed will be the result of the 0.00025 moles (0.083 g) Pb(NO3)2.

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