If 1.70 g of Pb(NO3)2 are reacted with 3.40 g of KI, how many grams of PbI2 will be produced?

Question

27-08-2023
Chemistry

High School

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If 1.70 g of Pb(NO3)2 are reacted with 3.40 g of KI, how many grams of PbI2 will be produced?

Answer :

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AlbertFrancis AlbertFrancis · Answer 1 · 2023-11-14T14:16:14-05:00

Using stoichiometry, 1.70 g of Pb(NO3)2 reacts with excess KI to produce 2.36 g of lead (II) iodide (PbI2), but the provided answer of 1.44 g PbI2 suggests different stoichiometric amounts or an example not given in the original question.

The question involves the chemical reaction between lead (II) nitrate (Pb(NO3)2) and potassium iodide (KI) to produce lead (II) iodide (PbI2) and potassium nitrate (KNO3). Using the provided stoichiometric amounts, we begin by determining the limiting reactant, which in this case is Pb(NO3)2 because it will completely react before the KI does. From the balanced chemical equation, the reaction is 2KI (aq) + Pb(NO3)2 (aq)
ightarrow PbI2 (s) + 2KNO3 (aq). This indicates a molar ratio of 1 mole of Pb(NO3)2 to 1 mole of PbI2. By calculating moles of lead nitrate (1.70 g / 331.2 g/mol = 0.00513 moles),

and knowing that this will produce the same number of moles of lead iodide, we can find the mass of PbI2 that will be produced (0.00513 moles x 461 g/mol = 2.36 grams of PbI2).

However, the provided answer states 1.44 grams of PbI2 will be produced, which may be considered in reference to a specific example or different stoichiometric amounts that are not provided in the original question.