Question

Calculate the mass of lead which would be obtained by heating 34.25 g of lead oxide (red lead, Pb₃O₄) in a stream of hydrogen and the mass of water formed at the same time. [Pb=207, H=1, O=16] Hint: I am giving the equation below for your convenience:

Pb₃O₄ + 4H₂ → 3Pb + 4H₂O

Answer 1

The mass of lead obtained is 26.54 g and the mass of water formed is 3.073 g.

Step-by-step explanation:

To calculate the mass of lead obtained and the mass of water formed, we need to use the equation provided:

Pb₃O₄ + 4H₂ → 3Pb + 4H₂O.

We know the molar mass of Pb₃O₄ is 207+207+207+16+16+16+16 = 802 g/mol.

We can calculate the number of moles of Pb₃O₄: 34.25 g ÷ 802 g/mol = 0.0427 mol.

According to the balanced equation, the mole ratio of Pb₃O₄ to Pb is 1:3. Therefore, we multiply the number of moles of Pb₃O₄ by 3 to get the moles of Pb: 0.0427 mol x 3 = 0.1281 mol.

The molar mass of Pb is 207 g/mol, so the mass of Pb obtained is: 0.1281 mol x 207 g/mol = 26.54 g.

Since the mole ratio of H₂O to Pb₃O₄ is 4:1, the moles of H₂O formed can be calculated by multiplying the moles of Pb₃O₄ by 4: 0.0427 mol x 4 = 0.1708 mol.

The molar mass of H₂O is 18 g/mol, so the mass of water formed is: 0.1708 mol x 18 g/mol = 3.073 g.