Answer to Question #309193 in General Chemistry for Nicole

Question

What mass of lead (II) iodide will be formed when 20.0 g of lead (II) nitrate is added to 33.0 g of potassium iodide?

Answer 1

The given chemical reaction can be represented as follows

$Pb(NO_3)_2 + 2KI \longrightarrow 2KNO_3 + PbI_2$

Mass of Lead nitrate =20 g,mass potassium iodide= 33 g

Mole of lead nitrate = =341. =0.06

Mole of potassium iodide =0.18

Mole of lead iodide = 0.06

Mass of lead iodide= 0.06×461= 27.66 g

So mass of lead iodide is 27.66g and 0.p6 mole