Lead (II) nitrate as Pb(NO 3)2 and potassium iodide as KI combine to form the products shown.

Question

2 KNO3 + PbI2
How many moles of each reactant are involved in the reaction?
(1 point)
3 2
3 2
1 mole of lead (II) nitrate and 2 moles of potassium iodide
2 moles of lead (II) nitrate and 2 moles of potassium iodide
1 mole of lead (II) nitrate and 1 mole of potassium iodide
2 moles of lead (II) nitrate and 1 mole of potassium iodide

Answer 1

To determine the moles of each reactant involved in the reaction, we should first look at the balanced chemical equation.

The unbalanced reaction is:

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

Balancing this reaction, we can write:

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

From this balanced equation, we can see that:

• 1 mole of lead (II) nitrate (Pb(NO₃)₂) reacts with 2 moles of potassium iodide (KI).

So the correct answer is: 1 mole of lead (II) nitrate and 2 moles of potassium iodide.