How many grams of PbBr2 will precipitate when excess CuBr2 solution is added to 77.0 mL of 0.595 M Pb (NO3) 2 solution? Pb (NO3) 2 (aq) + CuBr2 (aq) PbBr2 (s) + Cu (NO3) 2 (aq)

16.89g of PbBr2

Explanation:

First, let us calculate the number of mole of Pb (NO3) 2. This is illustrated below:

Molarity of Pb (NO3) 2 = 0.595M

Volume = 77mL = 77/1000 = 0.077L

Mole = ?

Molarity = mole/Volume

Mole = Molarity x Volume

Mole of Pb (NO3) 2 = 0.595x0.077

Mole of Pb (NO3) 2 = 0.046mol

Convert 0.046mol of Pb (NO3) 2 to grams as shown below:

Molar Mass of Pb (NO3) 2 =

207 + 2[ 14 + (16x3) ]

= 207 + 2[14 + 48]

= 207 + 2[62] = 207 + 124 = 331g/mol

Mass of Pb (NO3) 2 = number of mole x molar Mass = 0.046 x 331 = 15.23g

Molar Mass of PbBr2 = 207 + (2x80) = 207 + 160 = 367g/mol

Equation for the reaction is given below:

Pb (NO3) 2 + CuBr2 - > PbBr2 + Cu (NO3) 2

From the equation above,

331g of Pb (NO3) 2 precipitated 367g of PbBr2

Therefore, 15.23g of Pb (NO3) 2 will precipitate = (15.23x367) / 331 = 16.89g of PbBr2