Lead(II) nitrate, $$\(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\)$$, reacts with potassium iodide, KI, to form a yellow precipitate, $$\(\mathrm{PbI}_{2}\)$$, and a soluble salt, $$\(\mathrm{KNO}_{3}\)$$. What is the equation for the reaction?
A.
\(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{KI} \rightarrow \mathrm{PbI}_{2}+\mathrm{KNO}_{3}\)
B.
\(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{KI} \rightarrow \mathrm{PbI}_{2}+\mathrm{KNO}_{3}\)
C.
\(2 \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{KI} \rightarrow \mathrm{PbI}_{2}+2 \mathrm{KNO}_{3}\)
D.
\(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{KI} \rightarrow \mathrm{PbI}_{2}+2 \mathrm{KNO}_{3}\)
Exam No：0620_s20_qp_22 Year：2020 Question No：8
Answer:
D
Knowledge points:
3.1.8. Deduce the symbol equation with state symbols for a chemical reaction, given relevant information
Solution:
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