What mass precipitate will form of 1.50 L of highly concentrated Pb (Clo3) 2 is mixed with 0.300 L of 0.110 M NaI

Question

Reyna Beck

Chemistry

30 March, 17:30

What mass precipitate will form of 1.50 L of highly concentrated Pb (Clo3) 2 is mixed with 0.300 L of 0.110 M NaI

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Randall Diaz · Answer 1

Given:

1.50 L of highly concentrated Pb (Clo3) 2

0.300 L of 0.110 M NaI

Required:

mass precipitate of highly concentrated Pb (Clo3) 2

Solution:

M1V1 = M2V2

M1 (1.50 L of highly concentrated Pb (Clo3) 2) = (0.300 L) (0.110 M NaI)

M1 = 0.022M Pb (Clo3) 2 = 0.022 mol/L Pb (Clo3) 2

Molar mass of Pb (Clo3) 2 = 374.2 g/mol

Mass of Pb (Clo3) 2 = (0.022 mol/L Pb (Clo3) 2) (1.50 L) (374.2 g/mol) = 12.35g Pb (Clo3) 2