Question

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

Answer 1

(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}}=(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}}=(3.34g)/(207.2g/mole)=0.016mole`

The moles of Pb formed = 0.016 mole

Answer 2

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