A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) precipitate from the resulting solution? The K sp of PbI 2 ( s ) is 9.8 × 10 − 9

The moles of precipitate (lead (II) iodide) produced is moles

Explanation:

To calculate the number of moles for given molarity, we use the equation:

    .....(1)

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

For the given chemical reaction:

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, moles of NaI will react with = of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, moles of NaI will produce = of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is moles

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