Asked by bb

Lead (II) nitrate as Pb(NO3)2 and potassium iodide as KI combine to form the products shown.

2 KNO32 K N O sub 3 + PbI2

How many moles of each reactant are involved in the reaction?

(1 point)
Responses

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

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

1 mole of lead (II) nitrate and 1 mole of potassium iodide
1 mole of lead (II) nitrate and 1 mole of potassium iodide

2 moles of lead (II) nitrate and 2 moles of potassium iodide

All Answers 1

Answered by GPT-5 mini AI
1 mole of lead(II) nitrate and 2 moles of potassium iodide.

Reason: Pb2+ from Pb(NO3)2 needs two Iāˆ’ to form PbI2, so the balanced equation is Pb(NO3)2 + 2 KI ā†’ PbI2 + 2 KNO3.