Calculate the mass of lead that can be obtained by reducing lead (II) oxide with a mass of 260 g

Calculate the mass of lead that can be obtained by reducing lead (II) oxide with a mass of 260 g with carbon monoxide (II), if the yield is 82%.

1. The reduction of lead from its oxide by carbon monoxide occurs according to the equation:

PbO + CO = Pb + CO2 ↑;

2. Let’s calculate the chemical amount of lead oxide:

n (PbO) = m (PbO): M (PbO);

M (PbO) = 207 + 16 = 223 g / mol;

n (PbO) = 260: 223 = 1.1659 mol;

3. Let us determine the theoretical amount of reduced lead:

ntheor (Pb) = n (PbO) = 1.1659 mol;

4. Find the practical amount of lead:

npr (Pb) = ntheor (Pb) * ν = 1.1659 * 0.82 = 0.956 mol;

5. Let’s calculate the mass of lead:

m (Pb) = npr (Pb) * M (Pb) = 0.956 * 207 = 197.89 g.

Answer: 197.89 g.