Lead (II) nitrate as Pb(NO3)2 and potassium iodide as KI combine to form the products shown. 2 KNO3 + PbI2 How many moles of each reactant are involved in the reaction?

Question

A. 1 mole of lead (II) nitrate and 2 moles of potassium iodide
B. 1 mole of lead (II) nitrate and 1 mole of potassium iodide
C. 2 moles of lead (II) nitrate and 2 moles of potassium iodide
D. 2 moles of lead (II) nitrate and 1 mole of potassium iodide

Answer 1

To determine the correct stoichiometry of the reaction between lead (II) nitrate (Pb(NO3)2) and potassium iodide (KI) to yield potassium nitrate (KNO3) and lead (II) iodide (PbI2), we can start by writing a balanced chemical equation for the reaction.

The overall reaction can be represented as:

\[ \text{Pb(NO}_3\text{)}_2 + 2 \text{KI} \rightarrow 2 \text{KNO}_3 + \text{PbI}_2 \]

From this balanced equation, we can see that:

• 1 mole of lead (II) nitrate reacts with 2 moles of potassium iodide.

Thus, the correct answer is:

A. 1 mole of lead (II) nitrate and 2 moles of potassium iodide.