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Complete the Data Table:


Use these values:


The atomic mass of Al is 27.0 g/mole.


The atomic mass of Pb is 207.2 g/mole.


Use two significant digits for your answers:


Mass of Al wire before reaction = 3.96 g


Mass of Al wire after reaction = 3.65 g


Mass of Al lost_______ = g


Moles of Al lost =___________ moles


Mass of Pb + filter paper = 4.26 g


Mass of filter paper = 0.92 g


Mass of Pb =__________ g


Moles of Pb formed =___________ moles

Verified Answer :

syed514
  1. Mass of Al lost = 
    A. 0.31 g 

    2. Moles of Al lost= 0.31 g / 27.0 g/mol = 0.011 g Al 
    C. 0.011 moles 

    3. Mass of Pb=4.26 - 0.92 = 3.34 g 
    B. 3.34 g 

    4. Moles of Pb formed= 3.34 g / 207.2 g/mol = 0.016 mol Pb 
    A. 0.016

Answer :

(1) Mass of Al lost = 0.31 g

(2) The moles of Al lost = 0.0114 mole

(3) Mass of Pb = 3.34 g

(4) The moles of Pb formed = 0.016 mole

Explanation :

Solution for part (1) :

Mass of Al lost = Mass of Al wire before reaction - Mass of Al wire after reaction

Mass of Al lost = 3.96 g - 3.65 g = 0.31 g

Solution for part (2) :

\(\text{Moles of Al lost}=\frac{\text{Mass of Al lost}}{\text{Atomic mass of Al}}=\frac{0.31g}{27g/mole}=0.0114mole\)

The moles of Al lost = 0.0114 mole

Solution for part (3) :

As we know that,

Mass of Pb + Filter paper = 4.26 g

Mass of Pb = 4.26 g - Mass of filter paper

Mass of Pb = 4.26 g - 0.92 g

Mass of Pb = 3.34 g

Solution for part (4) :

\(\text{Moles of Pb formed}=\frac{\text{Mass of Pb formed}}{\text{Atomic mass of Pb}}=\frac{3.34g}{207.2g/mole}=0.016mole\)

The moles of Pb formed = 0.016 mole

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