how much pb can be isolated from 3.5kg of pbco3?
To determine how much Pb (lead) can be isolated from 3.5 kg of PbCO3 (lead carbonate), we need to calculate the molar mass of PbCO3 and then use stoichiometry to convert the mass of PbCO3 to the mass of Pb.
First, let's find the molar mass of PbCO3:
1. Molar mass of Pb = 207.2 g/mol (from periodic table)
2. Molar mass of C = 12.01 g/mol (from periodic table)
3. Molar mass of O = 16.00 g/mol (from periodic table)
Now, we can calculate the molar mass of PbCO3:
Molar mass of PbCO3 = (1 x Pb) + (1 x C) + (3 x O) = (1 x 207.2) + (1 x 12.01) + (3 x 16.00) = 267.21 g/mol
Next, we can use stoichiometry to determine the mass of Pb:
1 mole of PbCO3 yields 1 mole of Pb, according to the balanced chemical equation.
Since the molar mass of PbCO3 is 267.21 g/mol, we can set up a proportion to find the mass of Pb:
(1 mol PbCO3 / 267.21 g PbCO3) = (1 mol Pb / X g Pb)
Cross-multiplying the equation gives:
X g Pb = (1 mol Pb / 267.21 g PbCO3) * 3.5 kg PbCO3
Converting the mass of PbCO3 to grams:
3.5 kg PbCO3 = 3,500 g PbCO3
Plugging in the values:
X g Pb = (1 mol Pb / 267.21 g PbCO3) * 3,500 g PbCO3
Calculating:
X g Pb = (1/267.21) * 3,500 ≈ 13.10 g Pb
Therefore, approximately 13.10 grams of Pb can be isolated from 3.5 kg of PbCO3.