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26 December, 20:27

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)

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  1. 26 December, 20:39
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    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
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